{"id":626257,"date":"2023-09-09T10:50:20","date_gmt":"2023-09-09T10:50:20","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=626257"},"modified":"2023-09-09T10:50:32","modified_gmt":"2023-09-09T10:50:32","slug":"kc-sinha-exercise-21-1-mathematics-solution-class-11-chapter-21-dirghavrutta","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-21-1-mathematics-solution-class-11-chapter-21-dirghavrutta\/","title":{"rendered":"KC Sinha: Exercise 21.1- Mathematics Solution Class 11 Chapter 21 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924"},"content":{"rendered":"\n

<\/span><\/p>\n\n\n\n

\u0909\u0938 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u093e:
(i) \u090f\u0915 \u0928\u093e\u092d\u093f (-1,1) \u0928\u093f\u092f\u0924\u093e x-y+3=0 \u0924\u0925\u093e \u0909\u0924\u094d\u0915\u0947\u0928\u094d\u0926\u094d\u0930\u0924\u093e $\\frac{1}{2}$ \u0939\u0948 \u0964
<\/strong> Sol:<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0915\u0947\u0928\u094d\u0926\u094d\u0930 \u092e\u0942\u0932 \u092c\u093f\u0928\u094d\u0926\u0941 \u0939\u0948 \u0924\u0925\u093e \u091c\u093f\u0938\u0915\u093e \u0926\u0940\u0930\u094d\u0918 \u0905\u0915\u094d\u0937 x-\u0905\u0915\u094d\u0937 \u092a\u0930 \u0939\u0948 \u0964
2a=12
a=6 ,c=4
b2<\/sup>=
a2<\/sup>=62<\/sup>-42<\/sup>
=36-16
=20<\/p>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923<\/p>\n\n\n\n

$\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1$<\/p>\n\n\n\n

$\\frac{x^2}{6^2}+\\frac{y^2}{20}=1$<\/p>\n\n\n\n

$\\frac{x^2}{36}+\\frac{y^2}{20}=1$<\/p>\n\n\n\n

Question 4<\/h4>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u093e \u0915\u0947\u0928\u094d\u0926\u094d\u0930 (-2,3) \u0939\u0948, \u0924\u0925\u093e \u0905\u0930\u094d\u0927 \u0905\u0915\u094d\u0937\u094b \u0915\u0940 \u0932\u0902\u092c\u093e\u0907\u092f\u093e\u0901 3 \u0924\u0925\u093e 2 \u0939\u0948 \u091c\u092c \u0926\u0940\u0930\u094d\u0918 \u0905\u0915\u094d\u0937 y-\u0905\u0915\u094d\u0937 \u0915\u0947 \u0938\u092e\u093e\u0928\u094d\u0924\u0930 \u0939\u0948 \u0964
[Find the equation to the ellipse whose centre is (-2,3) and whose semi axes are 3 and 2 when the major axis is parallel to the y-axis]
<\/strong> Sol :
\u0909\u0938 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u091c\u093f\u0938\u0915\u093e \u0915\u0947\u0928\u094d\u0926\u094d\u0930 (?,\ua7b5) \u0939\u0948 \u0924\u0925\u093e \u0926\u0940\u0930\u094d\u0918 \u0905\u0915\u094d\u0937 y-\u0905\u0915\u094d\u0937 \u0915\u0947 \u0938\u092e\u093e\u0928\u0924\u0930 \u0939\u0948 \u0964<\/p>\n\n\n\n

$\\frac{(x-\\alpha)^2}{b^2}+\\frac{(y-\\beta)^2}{a^2}=1$<\/p>\n\n\n\n

\u2234$\\frac{(x+2)^2}{2^2}+\\frac{(y-3)^2}{3^2}=1$<\/p>\n\n\n\n

$\\frac{x^2+4x+4}{4}+\\frac{y^2-6x+9}{9}=1$<\/p>\n\n\n\n

$\\frac{9x^2+36x+36+4y^2-24y+36}{36}=1$<\/p>\n\n\n\n

9x2<\/sup>+4y2<\/sup>+36x-24y+72=36<\/p>\n\n\n\n

9x2<\/sup>+4y2<\/sup>+36x-24y+36=0<\/p>\n\n\n\n

Question 5<\/h4>\n\n\n\n

\u0909\u0938 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u0947 \u0936\u0940\u0930\u094d\u0937 (0,\u00b110) \u0924\u0925\u093e \u0909\u0924\u094d\u0915\u0947\u0928\u094d\u0926\u094d\u0930\u0924\u093e $\\frac{4}{5}$ \u0939\u0948 \u0964
[Find the equation of the eclipse having vertices (0,\u00b110) and eccentricity $\\frac{4}{5}$]
<\/strong> Sol :
\u0936\u0940\u0930\u094d\u0937 (0,\u00b110) ,$e=\\frac{4}{5}$<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0926\u0940\u0930\u094d\u0918\u0905\u0915\u094d\u0937 y-\u0905\u0915\u094d\u0937 \u092a\u0930 \u0939\u0948 \u0964
\u0924\u0925\u093e \u0915\u0947\u0928\u094d\u0926\u094d\u0930 \u092e\u0942\u0932 \u092c\u093f\u0928\u094d\u0926\u0941 \u0939\u0948 \u0964<\/p>\n\n\n\n

a=10 , $e=\\frac{4}{5}$<\/p>\n\n\n\n

b2<\/sup>=a2<\/sup>(1-e2<\/sup>)
b2<\/sup>=102<\/sup>
$\\left[1-\\left(\\frac{4}{5}\\right)^2\\right]$<\/p>\n\n\n\n

$b^2=100\\left(1-\\frac{16}{25}\\right)$<\/p>\n\n\n\n

$b^2=100\\left(\\frac{25-16}{25}\\right)$<\/p>\n\n\n\n

b2<\/sup>=4\u00d79<\/p>\n\n\n\n

b2<\/sup>=36
b=6<\/p>\n\n\n\n

\u2234\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923
$\\frac{x^2}{6^2}+\\frac{y^2}{10^2}=1$<\/p>\n\n\n\n

$\\frac{x^2}{36}+\\frac{y^2}{100}=1$<\/p>\n\n\n\n

$\\frac{25x^2+9y^2}{900}=1$<\/p>\n\n\n\n

25x2<\/sup>+9y2<\/sup>=900<\/p>\n\n\n\n

Question 6<\/h4>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u0932\u0947
[Find the equation of the ellipse]
<\/strong>
(i) \u091c\u093f\u0938\u0915\u093e \u0936\u0940\u0930\u094d\u0937 (0,\u00b13) \u0924\u0925\u093e \u0928\u093e\u092d\u093f\u092f\u093e\u0901 (0,\u00b15) \u092a\u0930 \u0939\u0948 \u0964
[having vertices at (0,\u00b113) and foci at (0,\u00b15)]
<\/strong> Sol :
\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0926\u0940\u0930\u094d\u0918 \u0905\u0915\u094d\u0937 y-\u0905\u0915\u094d\u0937 \u092a\u0930 \u0939\u094b\u0917\u093e \u0924\u0925\u093e \u0915\u0947\u0928\u094d\u0926\u094d\u0930 \u092e\u0942\u0932 \u092c\u093f\u0928\u094d\u0926\u0941 \u0939\u0948 \u0964
a=13 , c=5<\/p>\n\n\n\n

b2<\/sup>=a2<\/sup>-c2<\/sup>
b2<\/sup>=132<\/sup>-52<\/sup>
b=12<\/p>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923
$\\frac{x^2}{12^2}+\\frac{y^2}{13^2}=1$<\/p>\n\n\n\n

$\\frac{x^2}{144}+\\frac{y^2}{169}=1$<\/p>\n\n\n\n

(ii) \u091c\u093f\u0938\u0915\u093e \u0936\u0940\u0930\u094d\u0937 (\u00b15,0) \u0924\u0925\u093e \u0928\u093e\u092d\u093f\u092f\u093e\u0901 (\u00b14,0) \u0939\u0948 \u0964
[having vertices (\u00b15,0) and foci (\u00b14,0)]
<\/strong> Sol :
\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0936\u0940\u0930\u094d\u0937 \u0905\u0915\u094d\u0937 x-\u0905\u0915\u094d\u0937 \u092a\u0930 \u0939\u0948 \u0924\u0925\u093e \u0915\u0947\u0928\u094d\u0926\u094d\u0930 \u092e\u0942\u0932 \u092c\u093f\u0928\u094d\u0926\u0941 \u0939\u0948 \u0964
a=5 ,c=4<\/p>\n\n\n\n

b2<\/sup>=a2<\/sup>-c2<\/sup><\/p>\n\n\n\n

b2<\/sup>=52<\/sup>-42<\/sup><\/p>\n\n\n\n

b=3<\/p>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923<\/p>\n\n\n\n

$\\frac{x^2}{5^2}+\\frac{y^2}{3^2}=1$<\/p>\n\n\n\n

$\\frac{x^2}{25}+\\frac{y^2}{9}=1$<\/p>\n\n\n\n

Question 7<\/h4>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u0932\u0947 \u0964
[Find the equation of the ellipse having ]
<\/strong>
(i) \u091c\u093f\u0938\u0915\u0947 \u0926\u0940\u0930\u094d\u0918 \u0905\u0915\u094d\u0937 \u0915\u0940 \u0932\u092e\u094d\u092c\u093e\u0908 26 \u0924\u0925\u093e \u0928\u093e\u092d\u093f\u092f\u093e\u0901 (\u00b15,0) \u0939\u0948 \u0964
[Length of major axis 26 and foci (\u00b15,0)]
<\/strong> Sol :
\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0926\u0940\u0930\u094d\u0918 \u0905\u0915\u094d\u0937 x-\u0905\u0915\u094d\u0937 \u092a\u0930 \u0939\u0948 \u0924\u0925\u093e \u0915\u0947\u0928\u094d\u0926\u094d\u0930 \u092e\u0942\u0932 \u092c\u093f\u0928\u094d\u0926\u0941 \u0939\u0948 \u0964
2a=16
a=8<\/p>\n\n\n\n

b2<\/sup>=a2<\/sup>-c2<\/sup><\/p>\n\n\n\n

(iv) \u091c\u093f\u0938\u0915\u0947 \u0932\u0918\u0941 \u0905\u0915\u094d\u0937 \u0915\u0940 \u0932\u092e\u094d\u092c\u093e\u0908 16 \u0924\u0925\u093e \u0928\u093e\u092d\u093f\u092f\u093e\u0901 (0,\u00b16) \u0939\u0948 \u0964
[Length if minor axis 16 and foci (0,\u00b16)]
<\/strong> Sol :
c=6<\/p>\n\n\n\n

2b=16
b=8<\/p>\n\n\n\n

a2<\/sup>=b2<\/sup>+c2<\/sup><\/p>\n\n\n\n

Question 8<\/h4>\n\n\n\n

\u0909\u0938 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u094b \u092c\u093f\u0928\u094d\u0926\u0941 (3,2) \u0924\u0925\u093e (1,6) \u0938\u0947 \u0917\u0941\u091c\u0930\u0924\u093e \u0939\u0948 \u0924\u0925\u093e \u091c\u093f\u0938\u0915\u093e \u0915\u0947\u0928\u094d\u0926\u094d\u0930 (0,0) \u0939\u0948 \u0924\u0925\u093e \u0926\u0940\u0930\u094d\u0918 \u0905\u0915\u094d\u0937 , y-\u0905\u0915\u094d\u0937 \u092a\u0930 \u0939\u0948 \u0964
[Find the equation of the ellipse passing through the point (3,2) having centre at (0,0) and major axis on y-axis]
<\/strong> Sol :
\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u091c\u093f\u0938\u0915\u093e \u0926\u0940\u0930\u094d\u0918 \u0905\u0915\u094d\u0937 y-\u0905\u0915\u094d\u0937 \u092a\u0930 \u0939\u0948 \u0924\u0925\u093e \u0915\u0947\u0928\u094d\u0926\u094d\u0930 \u092e\u0942\u0932 \u092c\u093f\u0928\u094d\u0926\u0941 \u0939\u0948 \u0964
$\\frac{x^2}{b^2}+\\frac{y^2}{a^2}=1$<\/p>\n\n\n\n

\u2235\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 (3,2) \u0938\u0947 \u0917\u0941\u091c\u0930 \u0930\u0939\u093e \u0939\u0948 \u0964
$\\frac{3^2}{b^2}+\\frac{2^2}{a^2}=1$<\/p>\n\n\n\n

$\\frac{9}{b^2}+\\frac{4}{a^2}=1$..(i)<\/p>\n\n\n\n

\u2234\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 (1,6) \u0938\u0947 \u0917\u0941\u091c\u0930 \u0930\u0939\u093e \u0939\u0948 \u0964
$\\frac{1^2}{b^2}+\\frac{6^2}{a^2}=1$<\/p>\n\n\n\n

$\\frac{1}{b^2}+\\frac{36}{a^2}=1$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0924\u0925\u093e (ii) \u0938\u0947 ,<\/p>\n\n\n\n

$\\begin{aligned}\\frac{81}{b^2}+\\frac{36}{a^2}=9\\\\ \\frac{1}{b^2}+\\frac{36}{a^2}=1\\\\ \\hline \\frac{80}{b^2}=8 \\end{aligned}$
8b2<\/sup>=80
b2<\/sup>=10<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u0938\u0947,
$\\frac{1}{10}+\\frac{36}{a^2}=1$<\/p>\n\n\n\n

$\\frac{36}{a^2}=1-\\frac{1}{10}$<\/p>\n\n\n\n

$\\frac{36}{a^2}=\\frac{10-1}{10}$<\/p>\n\n\n\n

$\\frac{36}{a^2}=\\frac{9}{10}$
a2<\/sup>=40<\/p>\n\n\n\n

\u2234\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923
$\\frac{x^2}{10}+\\frac{y^2}{40}=1$<\/p>\n\n\n\n

Question 9<\/h4>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u0947 \u0926\u0940\u0930\u094d\u0918 \u0905\u0915\u094d\u0937 \u0915\u0947 \u091b\u094b\u0930 (0,\u00b1\u221a5) \u0924\u0925\u093e \u0905\u0915\u094d\u0937 \u0915\u0947 \u091b\u094b\u0930 (\u00b11,0) \u0939\u0948 \u0964
[Find the equation of the ellipse having ends of major axis (0,\u00b1\u221a5) and ends of minor axis  (\u00b11,0)]
<\/strong> Sol :
a=\u221a5 , b=1<\/p>\n\n\n\n

$\\frac{x^2}{b^2}+\\frac{y^2}{a^2}=1$<\/p>\n\n\n\n

Question 10<\/h4>\n\n\n\n

\u092f\u0926\u093f a \u0905\u0930\u094d\u0925 \u0926\u0940\u0930\u094d\u0918 \u0905\u0915\u094d\u0937 \u0915\u0940 \u0932\u092e\u094d\u092c\u093e\u0908 , b \u0905\u0930\u094d\u0927 \u0932\u0918\u0941 \u0905\u0915\u094d\u0937 \u0915\u0940 \u0932\u092e\u094d\u092c\u093e\u0908 \u0924\u0925\u093e c \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u0947 \u090f\u0915 \u0928\u093e\u092d\u093f \u0915\u0940 \u0915\u0947\u0928\u094d\u0926\u094d\u0930 \u0938\u0947 \u0926\u0942\u0930\u0940 \u0939\u0948 \u0924\u094b \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u0947 \u0932\u093f\u090f \u0915\u0947\u0928\u094d\u0926\u094d\u0930 (0,0) \u0928\u093e\u092d\u093f x-\u0905\u0915\u094d\u0937 \u092a\u0930, b=3 \u0924\u0925\u093e c=4 \u0939\u0948 \u0964
[If a be the length of semi major axis , b the length of semi minor axis and c the distance of one focus from the centre of an ellipse then find the equation of the ellipse for which centre is (0,0) foci is on x-axis , b=3 and c=4]
<\/strong> Sol :<\/p>\n\n\n\n

Question 11<\/h4>\n\n\n\n

\u090f\u0915 \u0926\u0940\u0930\u093f\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u0947 \u0928\u093e\u092d\u093f\u092f\u094b \u0915\u0947 \u092c\u0940\u091a \u0915\u0940 \u0926\u0942\u0930\u0940 10 \u0924\u0925\u093e \u0907\u0938\u0915\u0947 \u0928\u093e\u092d\u093f\u0932\u092e\u094d\u092c \u0915\u0940 \u0932\u092e\u094d\u092c\u093e\u0908 15 \u0939\u0948 , \u0924\u094b \u0907\u0938\u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u092c\u0915\u093f \u0907\u0938\u0915\u0947 \u0905\u0915\u094d\u0937 \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915 \u0905\u0915\u094d\u0937 \u0939\u0948 \u0964
[The distance between the foci of an ellipse is 10 and its latus rectum is 15, find its equation referred to its axes if coordinates]
<\/strong> Sol :
\u0928\u093e\u092d\u093f\u092f\u094b \u0915\u0947 \u092c\u0940\u091a \u0915\u0940 \u0926\u0942\u0930\u0940=10
2c=10
c=5<\/p>\n\n\n\n

\u0928\u093e\u092d\u093f\u0932\u0902\u092e\u094d\u092c \u0915\u0940 \u0932\u092e\u094d\u092c\u093e\u0908=15
$\\frac{2b^2}{a}=15$<\/p>\n\n\n\n

$\\frac{2(a^2-c^2)}{a}=15$<\/p>\n\n\n\n

2(a2<\/sup>-25)=15a
2a2<\/sup>-50=15a
2a2<\/sup>-15a-50=0
2a2<\/sup>-20a+5a-5a=0
2a(a-10)+5(a-10)=0
(2a+5)(a-10)=0<\/p>\n\n\n\n

$\\begin{array}{l|l}2a+5&a-10=0\\\\a=-\\frac{5}{2}&a=10\\end{array}$<\/p>\n\n\n\n

a=10
$\\frac{2b^2}{a}=15$<\/p>\n\n\n\n

$\\frac{2b^2}{10}=15$<\/p>\n\n\n\n

b2<\/sup>=75<\/p>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923
$\\frac{x^2}{10^2}+\\frac{y^2}{75}=1$<\/p>\n\n\n\n

$\\frac{x^2}{100}+\\frac{y^2}{75}=1$<\/p>\n\n\n\n

$\\frac{3x^2+4y^2}{300}=1$<\/p>\n\n\n\n

3x2<\/sup>+4y2<\/sup>=300<\/p>\n\n\n\n

Question 12<\/h4>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u092e\u093e\u0928\u0915 \u0930\u0941\u092a \u092e\u0947 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u093e \u0932\u0918\u0941 \u0905\u0915\u094d\u0937 \u0928\u093e\u092d\u093f\u092f\u094b \u0915\u0947 \u092c\u0940\u091a \u0915\u0940 \u0926\u0942\u0930\u0940 \u0915\u0947 \u092c\u0930\u093e\u092c\u0930 \u0939\u0948 \u0924\u0925\u093e \u091c\u093f\u0938\u0915\u093e \u0928\u093e\u092d\u093f\u0932\u092e\u094d\u092c 10 \u0939\u0948\u0964
[Find the equation of the ellipse in the standard form whose minor axis is equal tot he distance between the foci and whose latus rectum is 10]
<\/strong> Sol :
\u0932\u0918\u0941 \u0905\u0915\u094d\u0937 \u0915\u0940 \u0932\u0902\u092e\u094d\u092c\u093e\u0908=\u0928\u093e\u092d\u093f\u092f\u094b \u0915\u0947 \u092c\u0940\u091a \u0915\u0940 \u0926\u0942\u0930\u0940
2b=2c
b=c<\/p>\n\n\n\n

\u0928\u093e\u092d\u093f\u0932\u092e\u094d\u092c=10<\/p>\n\n\n\n

$\\frac{2b^2}{a}=10$<\/p>\n\n\n\n

\u2235a2<\/sup>=b2<\/sup>+c2<\/sup>
a2<\/sup>=b2<\/sup>+b2<\/sup>
a2<\/sup>=2b2<\/sup><\/p>\n\n\n\n

$\\frac{a^2}{a}=10$
a=10<\/p>\n\n\n\n

102<\/sup>=2b2<\/sup><\/p>\n\n\n\n

$\\frac{100}{2}=b^2$
b2<\/sup>=50
a2<\/sup>=50<\/p>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923
$\\frac{x^2}{100}+\\frac{y^2}{50}=1$<\/p>\n\n\n\n

$\\frac{x^2+2y^2}{100}=1$<\/p>\n\n\n\n

x2<\/sup>+2y2<\/sup>=100<\/p>\n\n\n\n

Question 13<\/h4>\n\n\n\n

\u0915\u093f\u0938\u0940 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u0940 \u0909\u0924\u094d\u0915\u0947\u0928\u094d\u0926\u094d\u0930\u0924\u093e $\\frac{1}{2}$ \u0939\u0948 \u0924\u0925\u093e \u0928\u093e\u092d\u093f\u092f\u094b \u0915\u0947 \u092c\u0940\u091a \u0915\u0940 \u0926\u0942\u0930\u0940 4 \u0907\u0915\u093e\u0908 \u0939\u0948 \u0964 \u092f\u0926\u093f \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0926\u0940\u0930\u094d\u0918 \u0905\u0915\u094d\u0937 \u0924\u0925\u093e \u0932\u0918\u0941 \u0905\u0915\u094d\u0937 \u0915\u094d\u0930\u092e\u0936\u0903 x-\u0905\u0915\u094d\u0937 \u0924\u0925\u093e y-\u0905\u0915\u094d\u0937 \u0915\u0947 \u0905\u0928\u0941\u0926\u093f\u0936 \u0939\u094b \u0924\u094b \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u0932\u0947 \u0964
<\/strong> Sol :
$e=\\frac{1}{2}$ ,<\/p>\n\n\n\n

2c=4
c=2<\/p>\n\n\n\n

$\\frac{c}{a}=\\frac{1}{2}$
$\\frac{2}{a}=\\frac{1}{2}$<\/p>\n\n\n\n

$b=\\sqrt{a^2-c^2}$
$=\\sqrt{4^2-2^2}$
$=\\sqrt{16-4}=\\sqrt{12}$<\/p>\n\n\n\n

$b=\\sqrt{12}$<\/p>\n\n\n\n

\u2234\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923
$\\frac{x^2}{4^2}+\\frac{y^2}{(\\sqrt{12})^2}=1$<\/p>\n\n\n\n

$\\frac{x^2}{16}+\\frac{y^2}{12}=1$<\/p>\n\n\n\n

$\\frac{3x^2+4y^2}{48}=1$<\/p>\n\n\n\n

3x2<\/sup>+4y2<\/sup>=48<\/p>\n\n\n\n

Question 14<\/h4>\n\n\n\n

(6,4) \u0938\u0947 \u091c\u093e\u0924\u0947 \u0939\u0941\u090f \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u0940 \u0928\u093e\u092d\u093f\u092f\u093e\u0901 y-\u0905\u0915\u094d\u0937 \u092a\u0930, \u0915\u0947\u0928\u094d\u0926\u094d\u0930 \u092e\u0942\u0932 \u092c\u093f\u0928\u094d\u0926\u0941 \u0924\u0925\u093e \u0909\u0924\u094d\u0915\u0947\u0928\u094d\u0926\u094d\u0930\u0924\u093e $\\frac{3}{4}$ \u0939\u0948 \u0964
[Find the equation of the ellipse passing through (6,4), foci on y-axis centre at the origin and having eccentricity $\\frac{3}{4}$]
<\/strong> Sol :
\u0930\u0947\u0916\u093e \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u091c\u093f\u0938\u0915\u0940 \u0928\u093e\u092d\u0940 y-\u0905\u0915\u094d\u0937 \u092a\u0930 \u0924\u0925\u093e \u0915\u0947\u0928\u094d\u0926\u094d\u0930 \u092e\u0942\u0932 \u092c\u093f\u0928\u094d\u0926\u0941 \u0939\u0948 \u0964
$\\frac{x^2}{b^2}+\\frac{y^2}{a^2}=1$<\/p>\n\n\n\n

\u2235\u0926\u0940\u0916\u0930\u094d\u0935\u0943\u0924\u094d\u0924 (6,4) \u0938\u0947 \u0917\u0941\u091c\u0930 \u0930\u0939\u093e \u0939\u0948 \u0964
$\\frac{6^2}{b^2}+\\frac{4^2}{a^2}=1$<\/p>\n\n\n\n

$\\frac{36}{b^2}+\\frac{4^2}{a^2}=1$..(i)<\/p>\n\n\n\n

\u2235$e=\\frac{3}{4}$<\/p>\n\n\n\n

$\\frac{c}{a}=\\frac{3}{4}$
$\\frac{c^2}{a^2}=\\frac{9}{16}$<\/p>\n\n\n\n

$\\frac{a^2-b^2}{a^2}=\\frac{9}{16}$<\/p>\n\n\n\n

16a2<\/sup>-16b2<\/sup>=9a2<\/sup>
7a2<\/sup>-16b2<\/sup>=0
7a2<\/sup>=16b2<\/sup><\/p>\n\n\n\n

$a^2=\\frac{16}{7}b^2$<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0938\u0947 ,<\/p>\n\n\n\n

$\\frac{36}{b^2}+\\frac{16}{}=1$<\/p>\n\n\n\n

$\\frac{36}{b^2}+\\dfrac{16}{\\frac{16}{7}b^2}=1$<\/p>\n\n\n\n

$\\frac{36}{b^2}+\\frac{7}{b^2}=1$<\/p>\n\n\n\n

$\\frac{43}{b^2}=1$<\/p>\n\n\n\n

b2<\/sup>=43<\/p>\n\n\n\n

\u2234$a^2=\\frac{16}{17}b^2$<\/p>\n\n\n\n

$=\\frac{16}{17}\\times 43$<\/p>\n\n\n\n

$a^2=\\frac{688}{7}$<\/p>\n\n\n\n

\u2234\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923<\/p>\n\n\n\n

$\\frac{x^2}{43}+\\dfrac{y^2}{\\frac{688}{7}}=1$<\/p>\n\n\n\n

$\\frac{x^2}{43}+\\frac{7y^2}{688}=1$<\/p>\n\n\n\n

$\\frac{16x^2+7y^2}{688}=1$<\/p>\n\n\n\n

16x2<\/sup>+7y2<\/sup>=688<\/p>\n\n\n\n

Question 15<\/h4>\n\n\n\n

(4,1) \u0938\u0947 \u091c\u093e\u0924\u0947 \u0939\u0941\u090f \u0926\u0940\u0930\u094d\u0916\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u0940 \u0928\u093e\u092d\u093f\u092f\u093e\u0901 (\u00b13,0) \u0939\u0948 \u0964
[Find the equation of the ellipse passing through (4,1) with foci as (\u00b13,0)]
<\/strong> Sol :
\u092e\u093e\u0928\u093e \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923
$\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1$<\/p>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 (4,1) \u0938\u0947 \u0917\u0941\u091c\u0930 \u0930\u0939\u093e \u0939\u0948 \u0964<\/p>\n\n\n\n

$\\frac{4^2}{a^2}+\\frac{1^2}{b^2}=1$<\/p>\n\n\n\n

$\\frac{16}{a^2}+\\frac{1^2}{b^2}=1$..(i)<\/p>\n\n\n\n

\u0928\u093e\u092d\u093f\u092f\u093e\u0901(\u00b13,0)
c=3
a2<\/sup>=b2<\/sup>+c2<\/sup>
a2<\/sup>=b2<\/sup>+32<\/sup><\/p>\n\n\n\n

a2<\/sup>=b2<\/sup>+9<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0938\u0947
$\\frac{16}{a^2}+\\frac{1}{b^2}=1$<\/p>\n\n\n\n

$\\frac{16}{b^2+9}+\\frac{1}{b^2}=1$<\/p>\n\n\n\n

$\\frac{16b^2+b^2+9}{b^2(b^2+9)}=1$<\/p>\n\n\n\n

$\\frac{17b^2+9}{b^4+9b^2}=1$<\/p>\n\n\n\n

b4<\/sup>+9b2<\/sup>=17b2<\/sup>+9<\/p>\n\n\n\n

b4<\/sup>+9b2<\/sup>-17b2<\/sup>-9=0
b4<\/sup>-8b2<\/sup>-9=0
b4<\/sup>-9b2<\/sup>+b2<\/sup>-9=0<\/p>\n\n\n\n

b2<\/sup>(b2<\/sup>-9)+1(b2<\/sup>-9)=0<\/p>\n\n\n\n

(b2<\/sup>-9)(b2<\/sup>+1)=0<\/p>\n\n\n\n

$\\begin{array}{l|l}b^2-9=0& b^2+1=0\\\\b^2=9&b^2=-1\\end{array}$<\/p>\n\n\n\n

\u2235a2<\/sup>=b2<\/sup>+9<\/p>\n\n\n\n

a2<\/sup>=9+9<\/p>\n\n\n\n

=18<\/p>\n\n\n\n

\u2234\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 <\/p>\n\n\n\n

$\\frac{x^2}{18}+\\frac{y^2}{9}=1$<\/p>\n\n\n\n

Question 16<\/h4>\n\n\n\n

\u0909\u0928 \u0938\u092d\u0940 \u092c\u093f\u0928\u094d\u0926\u0941\u0913 \u0915\u0947 \u0938\u092e\u0941\u091a\u094d\u091a\u092f \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u0940 (0,4) \u0938\u0947 \u0926\u0942\u0930\u093f\u092f\u093e\u0901 \u0909\u0928\u0915\u0940 \u0930\u0947\u0916\u093e y=9 \u0938\u0947 \u0926\u0942\u0930\u093f\u092f\u094b \u0915\u093e $\\frac{2}{3}$ \u0939\u0948\u0964
[Find the equation of the set of all points whose distances from (0,4) are $\\frac{2}{3}$ of their distances from the line y=9]<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

\u092e\u093e\u0928\u093e \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0909\u0928 \u0938\u092d\u0940 \u092c\u093f\u0928\u094d\u0926\u0941\u0913 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u0939\u0948 \u0964
\u092e\u093e\u0928\u093e P(x,y) \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u092a\u0930 \u0938\u094d\u0925\u093f\u0924 \u092c\u093f\u0928\u094d\u0926\u0941 \u0939\u0948 \u0964<\/p>\n\n\n\n

$PS=\\frac{2}{3}\\times PM$<\/p>\n\n\n\n

$\\left| \\sqrt{(x-0)^2+(y-4)^2}=\\frac{2}{3}\\times \\left$|\\frac{y-9}{\\sqrt{0^2+1^2}}\\right|$<\/p>\n\n\n\n

$\\sqrt{x^2+(y-4)^2}=\\frac{2}{3}\\left|y-9\\right|$<\/p>\n\n\n\n

\u0926\u094b\u0928\u094b \u0924\u0930\u092b \u0935\u0930\u094d\u0917 \u0915\u0930\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

x2<\/sup>+(y-4)2<\/sup>=$\\frac{4}{9}(y-9)^2$<\/p>\n\n\n\n

x2<\/sup>+y2<\/sup>-8y+16=$\\frac{4}{9}(y^2-18y+81)$<\/p>\n\n\n\n

9x2<\/sup>+9y2<\/sup>-72y+144=4y2<\/sup>-72y+324<\/p>\n\n\n\n

9x2<\/sup>+5y2<\/sup>=324-144<\/p>\n\n\n\n

9x2<\/sup>+5y2<\/sup>=180<\/p>\n\n\n\n

Question 17<\/h4>\n\n\n\n

\u0909\u0938 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u0940 \u0928\u093e\u092d\u093f\u092f\u093e\u0901 (4,0) \u0924\u0925\u093e (-4,0) \u0924\u0925\u093e \u0909\u0924\u094d\u0915\u0947\u0928\u094d\u0926\u094d\u0930\u0924\u093e $\\frac{1}{3}$ \u0939\u0948 \u0964
[Find the equation to the ellipse whose foci are (4,0) and (-4,0) and eccentricity is $\\frac{1}{3}$]
<\/strong> Sol :<\/p>\n\n\n\n

Question 19<\/h4>\n\n\n\n

\u092f\u0926\u093f \u0915\u093f\u0938\u0940 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u0947 \u0928\u093e\u092d\u093f\u092f\u094b \u0915\u094b \u0907\u0938\u0915\u0947 \u0932\u0918\u0941 \u0905\u0915\u094d\u0937 \u0915\u0947 \u090f\u0915 \u091b\u094b\u0930 \u0938\u0947 \u092e\u093f\u0932\u093e\u0928\u0947 \u0935\u093e\u0932\u0940 \u0930\u0947\u0916\u093e\u0913 \u0915\u0947 \u092c\u0940\u091a \u0915\u093e \u0915\u094b\u0923 90\u00b0 \u0939\u0948 , \u0924\u094b \u0907\u0938\u0915\u093e \u0909\u0924\u094d\u0915\u0947\u0928\u094d\u0926\u094d\u0930\u0924\u093e \u0928\u093f\u0915\u093e\u0932\u0947 \u0964 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u092d\u0940 \u0928\u093f\u0915\u093e\u0932\u0947 \u092f\u0926\u093f \u0907\u0938\u0915\u093e \u0926\u0940\u0930\u094d\u0918 \u0905\u0915\u094d\u0937 2\u221a2 \u0939\u0948 \u0964
[If the angle between the line joining the foci of any ellipse to an extremity of the minor axis is 90\u00b0 , find the eccentricity. Find also the equation of the ellipse if the major axis is 2\u221a2]
<\/strong> Sol :
\u092e\u093e\u0928\u093e \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u0940 \u0928\u093e\u092d\u093f\u092f\u093e\u0901 S’ \u0924\u0925\u093e S \u0939\u0948 \u0964<\/p>\n\n\n\n

\u2220SBS’=90\u00b0<\/p>\n\n\n\n

\u2220S’BD=$\\frac{1}{2}\\times 90$
=45\u00b0<\/p>\n\n\n\n

tan45\u00b0=$\\frac{c}{b}$
$1=\\frac{c}{b}$
b=c<\/p>\n\n\n\n

a2<\/sup>=b2<\/sup>+c2<\/sup>
a2<\/sup>=b2<\/sup>+b2<\/sup>
a2<\/sup>=2b2<\/sup><\/p>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918 \u0905\u0915\u094d\u0937 \u0915\u0940 \u0932\u0902\u092c\u093e\u0908=2\u221a2
2a=2\u221a2<\/p>\n\n\n\n

2=2b2<\/sup>
b=\u00b11<\/p>\n\n\n\n

b=1 , c=1<\/p>\n\n\n\n

$e=\\frac{c}{a}=\\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923
$\\frac{x^2}{(\\sqrt{2})^2}+\\frac{y^2}{1^2}=1$<\/p>\n\n\n\n

Question 20<\/h4>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 9x2<\/sup>+16y2<\/sup>=144 \u0915\u0947 \u0932\u093f\u090f \u0926\u0940\u0930\u094d\u0918 \u0905\u0915\u094d\u0937 \u0924\u0925\u093e \u0932\u0918\u0941 \u0905\u0915\u094d\u0937 \u0915\u0940 \u0932\u092e\u094d\u092c\u093e\u0907\u092f\u093e\u0901 , \u0909\u0924\u094d\u0915\u0947\u0928\u094d\u0926\u094d\u0930\u0924\u093e, \u0928\u093e\u092d\u093f\u092f\u094b \u0915\u0947 \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915 \u0936\u0940\u0930\u094d\u0937 \u0924\u0925\u093e \u0928\u093f\u092f\u0924\u093e\u0913 \u0915\u0947 \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u0932\u0947 \u0964
[For the ellipse 9x2<\/sup>+16y2<\/sup>=144 , find the length of the major and minor axes, the eccentricity , the coordinates of the foci, the vertices and the equations of the directrices]
<\/strong> Sol :
\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923
9x2<\/sup>+16y2<\/sup>=144<\/p>\n\n\n\n

$\\frac{9x^2}{144}+\\frac{16y^2}{144}=1$<\/p>\n\n\n\n

$\\frac{x^2}{4^2}+\\frac{y^2}{3^2}=1$<\/p>\n\n\n\n

\u0905\u0924\u0903 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0926\u0940\u0930\u094d\u0918\u0905\u0915\u094d\u0937 x-\u0905\u0915\u094d\u0937 \u0939\u094b\u0917\u093e \u0924\u0925\u093e \u0915\u0947\u0928\u094d\u0926\u094d\u0930 , \u092e\u0942\u0932 \u092c\u093f\u0928\u094d\u0926\u0941 \u0939\u094b\u0917\u093e \u0964
a=4 ,b=3 ,
$c=\\sqrt{a^2-b^2}$
$c=\\sqrt{4^2-3^2}=\\sqrt{7}$<\/p>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918 \u0905\u0915\u094d\u0937 \u0915\u0940 \u0932\u0902\u092e\u094d\u092c\u093e\u0908=2a
=2\u00d74=8<\/p>\n\n\n\n

\u0932\u0918\u0941 \u0905\u0915\u094d\u0937 \u0915\u0940 \u0932\u092e\u094d\u092c\u093e\u0908=2b
=2\u00d73=6<\/p>\n\n\n\n

$e=\\frac{c}{a}=\\frac{\\sqrt{7}}{4}$
\u0928\u093e\u092d\u093f(\u00b1\u221a7,0)<\/p>\n\n\n\n

\u0936\u0940\u0930\u094d\u0937 (\u00b14,0)<\/p>\n\n\n\n

\u0928\u093f\u092f\u0924\u093e \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923
$x=\\pm \\frac{a}{e}$<\/p>\n\n\n\n

$x=\\pm \\dfrac{4}{\\frac{\\sqrt{7}}{4}}$<\/p>\n\n\n\n

$x=\\pm \\frac{16}{\\sqrt{7}}$<\/p>\n\n\n\n

Question 23<\/h4>\n\n\n\n

\u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0938\u092e\u0940\u0915\u0930\u0923 \u090f\u0915 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u094b \u0928\u093f\u0930\u0942\u092a\u093f\u0924 \u0915\u0930\u0924\u093e \u0939\u0948 \u0964 \u0907\u0938\u0915\u093e \u0915\u0947\u0928\u094d\u0926\u094d\u0930 \u0914\u0930 \u0909\u0924\u094d\u0915\u0947\u0928\u094d\u0926\u094d\u0930\u0924\u093e \u0928\u093f\u0915\u093e\u0932\u0947 \u0964
[Show that the following equation represents an ellipse and find its centre and eccentricity]
8x2<\/sup>+6y2<\/sup>-6x+12y+13=0
<\/strong> Sol :
\u21d28x2<\/sup>-6y2<\/sup>+6x+12y=-13<\/p>\n\n\n\n

\u21d28[x2<\/sup>-2x]+6[y2<\/sup>-2y]=-13<\/p>\n\n\n\n

\u21d28[x2<\/sup>-2.x.1+12<\/sup>]+6[y2<\/sup>+2.y.1+12<\/sup>-12<\/sup>]=-13<\/p>\n\n\n\n

\u21d28(x2<\/sup>-1)-8+6(y2<\/sup>+1)-6=-13<\/p>\n\n\n\n

\u21d28(x-1)2<\/sup>+6[y-(-1)]2<\/sup>=11<\/p>\n\n\n\n

\u21d2$\\dfrac{(x-1)^2}{\\frac{1}{8}}+\\dfrac{[y-(-1)]^2}{\\frac{1}{6}}=1$<\/p>\n\n\n\n

\u21d2$\\dfrac{(x-1)^2}{\\frac{1}{\\left(\\sqrt{8}\\right)^2}}+\\dfrac{[y-(-1)]^2}{\\frac{1}{\\left(\\sqrt{6}}\\right)}=1$<\/p>\n\n\n\n

$a=\\frac{1}{\\sqrt{6}},b=\\frac{1}{\\sqrt{8}}$<\/p>\n\n\n\n

\u0915\u0947\u0928\u094d\u0926\u094d\u0930 (1,-1)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u094b \u0928\u093f\u0930\u0941\u092a\u0940\u0924 \u0915\u0930\u0924\u093e \u0939\u0948 \u0964<\/p>\n\n\n\n

$C=\\sqrt{a^2-b^2}$<\/p>\n\n\n\n

$=\\sqrt{\\frac{1}{6}-\\frac{1}{8}}$<\/p>\n\n\n\n

$=\\sqrt{\\frac{4-3}{24}}=\\sqrt{\\frac{1}{24}}$<\/p>\n\n\n\n

$e=\\frac{c}{a}=\\dfrac{\\frac{1}{\\sqrt{24}}}{\\frac{1}{\\sqrt{6}}}$
$=\\frac{1}{2}$<\/p>\n\n\n\n

Question 24<\/h4>\n\n\n\n

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0915\u0947\u0928\u094d\u0926\u094d\u0930 \u0905\u0915\u094d\u0937\u094b \u0915\u0940 \u0932\u092e\u094d\u092c\u093e\u0908\u092f\u093e\u0901 \u0924\u0925\u093e \u0909\u0924\u094d\u0915\u0947\u0928\u094d\u0926\u094d\u0930\u0924\u093e \u0928\u093f\u0915\u093e\u0932\u0947 \u0964
[Find the centre, the lengths of the axes and the eccentricity of the ellipse ]
2x2<\/sup>+3y2<\/sup>-4x-12y+13=0
<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u21d22x2<\/sup>+3y2<\/sup>-4x-12y=-13<\/p>\n\n\n\n

\u21d22[x2<\/sup>-2x]+3(y2<\/sup>-4y)=-13<\/p>\n\n\n\n

\u21d22[x2<\/sup>-2.x.1+12<\/sup>-12<\/sup>]+3[y2<\/sup>-2.y.2+22<\/sup>-22<\/sup>]=-13<\/p>\n\n\n\n

\u21d22(x2<\/sup>-1)2<\/sup>-2+3(y-2)2<\/sup>-12=-13<\/p>\n\n\n\n

\u21d22(x-1)2<\/sup>+3(y-2)2<\/sup>=1<\/p>\n\n\n\n

\u21d2$\\dfrac{(x-1)^2}{\\frac{1}{2}}+\\dfrac{(y-2)^2}{\\frac{1}{3}}=1$<\/p>\n\n\n\n

\u21d2$\\dfrac{(x-1)^2}{\\left(\\frac{1}{2}\\right)^2}+\\dfrac{(y-2)^2}{\\left(\\frac{1}{3}\\right)^2}=1$<\/p>\n\n\n\n

\u21d2$a=\\frac{1}{\\sqrt{2}}$ ,$b=\\frac{1}{\\sqrt{3}}$<\/p>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 x-\u0905\u0915\u094d\u0937 \u0915\u0947 \u0905\u0928\u0941\u0926\u0940\u0936 \u0939\u0948\u0964<\/p>\n\n\n\n

\u0915\u0947\u0928\u094d\u0926\u094d\u0930 (1,2)<\/p>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918 \u0905\u0915\u094d\u0937 \u0915\u093e \u0932\u092e\u094d\u092c\u093e\u0908 $=2\\times \\frac{1}{\\sqrt{2}}=\\sqrt{2}$<\/p>\n\n\n\n

\u0932\u0918\u0941 \u0905\u0915\u094d\u0937 \u0915\u0940 \u0932\u092e\u094d\u092c\u093e\u0908 $=2b=2\\times \\frac{1}{\\sqrt{3}}=\\frac{2}{\\sqrt{3}}$<\/p>\n\n\n\n

$c=\\sqrt{\\left(\\frac{1}{\\sqrt{2}}\\right)^2-\\left(\\frac{1}{\\sqrt{3}}\\right)^2}$<\/p>\n\n\n\n

$=\\frac{1}{2}-\\frac{1}{3}$<\/p>\n\n\n\n

$e=\\frac{c}{a}=\\dfrac{\\frac{1}{\\sqrt{6}}}{\\frac{1}{\\sqrt{2}}}$<\/p>\n\n\n\n

$=\\frac{1}{\\sqrt{3}}$<\/p>\n\n\n\n

Question 27<\/h4>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0909\u0924\u094d\u0915\u0947\u0928\u094d\u0926\u094d\u0930\u0924\u093e \u0928\u093f\u0915\u093e\u0932\u0947 \u092f\u0926\u093f \u0907\u0938\u0915\u093e \u0928\u093e\u092d\u093f\u0932\u0902\u092c \u0907\u0938\u0915\u0947 \u0926\u0940\u0930\u094d\u0918 \u0905\u0915\u094d\u0937 \u0915\u093e \u0906\u0927\u093e \u0939\u0948 \u0964
[Find the eccentricity of an ellipse if its latus rectum is equal to one-half of its major axis]
<\/strong> Sol :
\u0928\u093e\u092d\u093f\u0932\u0902\u092c \u0915\u0940 \u0932\u092e\u094d\u092c\u093e\u0908$=\\frac{1}{2}\\times $\u0926\u0940\u0930\u094d\u0918 \u0905\u0915\u094d\u0937 \u0915\u0940 \u0932\u0902\u092e\u094d\u092c\u093e\u0908<\/p>\n\n\n\n

$\\frac{2b^2}{a}=\\frac{1}{2}\\times 2a$<\/p>\n\n\n\n

2b2<\/sup>=a2<\/sup><\/p>\n\n\n\n

$b^2=\\frac{a^2}{2}$<\/p>\n\n\n\n

\u2235a2<\/sup>=b2<\/sup>+c2<\/sup><\/p>\n\n\n\n

$a^2=\\frac{a^2}{2}+c^2$<\/p>\n\n\n\n

$c^2=a^2-\\frac{a^2}{2}$<\/p>\n\n\n\n

$c^2=\\frac{2a^2-a^2}{2}$<\/p>\n\n\n\n

$c^2=\\frac{a^2}{2}$<\/p>\n\n\n\n

$\\frac{c^2}{a^2}=\\frac{1}{2}$<\/p>\n\n\n\n

$\\left(\\frac{c}{a}\\right)^2=\\frac{1}{2}$<\/p>\n\n\n\n

$e=\\sqrt{\\frac{1}{2}}$<\/p>\n\n\n\n

$e=\\frac{1}{\\sqrt{2}}$<\/p>\n\n\n\n

Question 29<\/h4>\n\n\n\n

15 \u0938\u0947\u092e\u0940 \u0932\u0902\u092c\u0940 \u090f\u0915 \u091b\u0921\u093c AB \u0926\u094b\u0928\u094b \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915\u094b \u0915\u0947 \u092c\u0940\u091a \u092e\u0947 \u0907\u0938 \u092a\u094d\u0930\u0915\u093e\u0930 \u0930\u0916\u0940 \u0917\u092f\u0940 \u0939\u0948 \u0915\u093f \u0909\u0938\u0915\u093e \u090f\u0915 \u0938\u093f\u0930\u093e A, x-\u0905\u0915\u094d\u0937 \u092a\u0930 \u0914\u0930 \u0926\u0942\u0938\u0930\u093e \u0938\u093f\u0930\u093e B, y-\u0905\u0915\u094d\u0937 \u092a\u0930 \u0930\u0939\u0924\u093e \u0939\u0948 \u091b\u0921\u093c \u092a\u0930 \u090f\u0915 \u092c\u093f\u0928\u094d\u0926\u0941 P(x,y) \u0907\u0938 \u092a\u094d\u0930\u0915\u093e\u0930 \u0932\u093f\u092f\u093e \u0917\u092f\u093e \u0939\u0948 \u0915\u093f AP=6 \u0938\u0947\u092e\u0940 \u0939\u0948 \u0964 \u0926\u093f\u0916\u093e\u0907\u090f \u0915\u093f P \u0915\u093e \u092c\u093f\u0928\u094d\u0926\u0941\u092a\u0925 \u090f\u0915 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0939\u0948 \u0964
<\/strong> Sol :<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

\u092e\u093e\u0928\u093e AB \u091b\u0921\u093c \u0907\u0938 \u092a\u094d\u0930\u0915\u093e\u0930 \u0938\u094d\u0925\u093f\u0924 \u0939\u0948 \u0964 \u0915\u093f A \u0938\u0947 \u092e\u0942\u0932 \u092c\u093f\u0928\u094d\u0926\u0941 \u0915\u0940 \u0926\u0942\u0930\u0940 a \u0939\u0948 \u0924\u0925\u093e B \u0938\u0947 \u092e\u0942\u0932 \u092c\u093f\u0928\u094d\u0926\u0941 \u0915\u0940 \u0926\u0942\u0930\u0940 b \u0939\u0948 \u0964<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e P(x,y) ,AB \u092a\u0930 \u0938\u094d\u0925\u093f\u0924 \u0939\u0948 \u0964
AP=6cm ,<\/p>\n\n\n\n

BP=15-6
=9cm<\/p>\n\n\n\n

$\\frac{AP}{BP}=\\frac{6}{8}$
=2:3=m:n<\/p>\n\n\n\n

P(x,y)=$\\left(\\frac{2\\times 0+3\\times a}{2+3},\\frac{2\\times b+3\\times 0}{2+3}\\right)$<\/p>\n\n\n\n

$=\\left(\\frac{3a}{5},\\frac{2b}{5}\\right)$<\/p>\n\n\n\n

$x=\\frac{3a}{5}$,$y=\\frac{2b}{5}$
$a=\\frac{5x}{3}$,$b=\\frac{5y}{2}$<\/p>\n\n\n\n

OA2<\/sup>+OB2<\/sup>=AB2<\/sup><\/p>\n\n\n\n

$\\left(\\frac{5x}{3}\\right)^2+\\left(\\frac{5y}{2}\\right)^2=(15)^2$<\/p>\n\n\n\n

$\\frac{25x^2}{9}+\\frac{25y^2}{4}=225$<\/p>\n\n\n\n

$\\frac{25x^2}{9\\times 225}+\\frac{25y^2}{4\\times 225}=1$<\/p>\n\n\n\n

$\\frac{x^2}{81}+\\frac{y^2}{36}=1$<\/p>\n\n\n\n

$\\frac{x^2}{9^2}+\\frac{y^2}{6^2}=1$<\/p>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 11 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

\u0909\u0938 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u093e:(i) \u090f\u0915 \u0928\u093e\u092d\u093f (-1,1) \u0928\u093f\u092f\u0924\u093e x-y+3=0 \u0924\u0925\u093e \u0909\u0924\u094d\u0915\u0947\u0928\u094d\u0926\u094d\u0930\u0924\u093e $\\frac{1}{2}$ \u0939\u0948 \u0964 Sol: \u092e\u093e\u0928\u093e \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0915\u0947\u0928\u094d\u0926\u094d\u0930 \u092e\u0942\u0932 \u092c\u093f\u0928\u094d\u0926\u0941 \u0939\u0948 \u0924\u0925\u093e \u091c\u093f\u0938\u0915\u093e \u0926\u0940\u0930\u094d\u0918 \u0905\u0915\u094d\u0937 x-\u0905\u0915\u094d\u0937 \u092a\u0930 \u0939\u0948 \u09642a=12a=6 ,c=4b2=a2=62-42=36-16=20 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 $\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1$ $\\frac{x^2}{6^2}+\\frac{y^2}{20}=1$ $\\frac{x^2}{36}+\\frac{y^2}{20}=1$ Question 4 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u093e \u0915\u0947\u0928\u094d\u0926\u094d\u0930 (-2,3) \u0939\u0948, \u0924\u0925\u093e \u0905\u0930\u094d\u0927 \u0905\u0915\u094d\u0937\u094b \u0915\u0940 \u0932\u0902\u092c\u093e\u0907\u092f\u093e\u0901 3 […]<\/p>\n","protected":false},"author":302,"featured_media":626291,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[919],"tags":[],"boards":[],"class_list":["post-626257","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-11","entry"],"yoast_head":"\nKC Sinha: Exercise 21.1- Mathematics Solution Class 11 Chapter 21 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"\u0909\u0938 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u093e:(i) \u090f\u0915 \u0928\u093e\u092d\u093f (-1,1) \u0928\u093f\u092f\u0924\u093e x-y+3=0 \u0924\u0925\u093e \u0909\u0924\u094d\u0915\u0947\u0928\u094d\u0926\u094d\u0930\u0924\u093e $frac{1}{2}$ \u0939\u0948 \u0964 Sol: \u092e\u093e\u0928\u093e \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0915\u0947\u0928\u094d\u0926\u094d\u0930 \u092e\u0942\u0932 \u092c\u093f\u0928\u094d\u0926\u0941 \u0939\u0948\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-21-1-mathematics-solution-class-11-chapter-21-dirghavrutta\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"KC Sinha: Exercise 21.1- Mathematics Solution Class 11 Chapter 21 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924\" \/>\n<meta property=\"og:description\" content=\"\u0909\u0938 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u093e:(i) \u090f\u0915 \u0928\u093e\u092d\u093f (-1,1) \u0928\u093f\u092f\u0924\u093e x-y+3=0 \u0924\u0925\u093e \u0909\u0924\u094d\u0915\u0947\u0928\u094d\u0926\u094d\u0930\u0924\u093e $frac{1}{2}$ \u0939\u0948 \u0964 Sol: \u092e\u093e\u0928\u093e \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0915\u0947\u0928\u094d\u0926\u094d\u0930 \u092e\u0942\u0932 \u092c\u093f\u0928\u094d\u0926\u0941 \u0939\u0948\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-21-1-mathematics-solution-class-11-chapter-21-dirghavrutta\/\" \/>\n<meta property=\"og:site_name\" content=\"IndCareer Schools\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/indcareer\" \/>\n<meta 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\u0939\u0948","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-21-1-mathematics-solution-class-11-chapter-21-dirghavrutta\/","og_locale":"en_US","og_type":"article","og_title":"KC Sinha: Exercise 21.1- Mathematics Solution Class 11 Chapter 21 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924","og_description":"\u0909\u0938 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u093e:(i) \u090f\u0915 \u0928\u093e\u092d\u093f (-1,1) \u0928\u093f\u092f\u0924\u093e x-y+3=0 \u0924\u0925\u093e \u0909\u0924\u094d\u0915\u0947\u0928\u094d\u0926\u094d\u0930\u0924\u093e $frac{1}{2}$ \u0939\u0948 \u0964 Sol: \u092e\u093e\u0928\u093e \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0915\u0947\u0928\u094d\u0926\u094d\u0930 \u092e\u0942\u0932 \u092c\u093f\u0928\u094d\u0926\u0941 \u0939\u0948","og_url":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-21-1-mathematics-solution-class-11-chapter-21-dirghavrutta\/","og_site_name":"IndCareer Schools","article_publisher":"https:\/\/www.facebook.com\/indcareer","article_published_time":"2023-09-09T10:50:20+00:00","article_modified_time":"2023-09-09T10:50:32+00:00","og_image":[{"width":1600,"height":901,"url":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-22-1-scaled.jpg","type":"image\/jpeg"}],"author":"Pooja","twitter_card":"summary_large_image","twitter_creator":"@indcareer","twitter_site":"@indcareer","twitter_misc":{"Written by":"Pooja","Est. reading time":"9 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-21-1-mathematics-solution-class-11-chapter-21-dirghavrutta\/#article","isPartOf":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-21-1-mathematics-solution-class-11-chapter-21-dirghavrutta\/"},"author":{"name":"Pooja","@id":"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e"},"headline":"KC Sinha: Exercise 21.1- Mathematics Solution Class 11 Chapter 21 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Exercise 21.1- Mathematics Solution Class 11 Chapter 21 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 - IndCareer Schools","isPartOf":{"@id":"https:\/\/www.indcareer.com\/schools\/#website"},"primaryImageOfPage":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-21-1-mathematics-solution-class-11-chapter-21-dirghavrutta\/#primaryimage"},"image":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-21-1-mathematics-solution-class-11-chapter-21-dirghavrutta\/#primaryimage"},"thumbnailUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-22-1-scaled.jpg","datePublished":"2023-09-09T10:50:20+00:00","dateModified":"2023-09-09T10:50:32+00:00","description":"\u0909\u0938 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u093e:(i) \u090f\u0915 \u0928\u093e\u092d\u093f (-1,1) \u0928\u093f\u092f\u0924\u093e x-y+3=0 \u0924\u0925\u093e \u0909\u0924\u094d\u0915\u0947\u0928\u094d\u0926\u094d\u0930\u0924\u093e $\\frac{1}{2}$ \u0939\u0948 \u0964 Sol: \u092e\u093e\u0928\u093e \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0915\u0947\u0928\u094d\u0926\u094d\u0930 \u092e\u0942\u0932 \u092c\u093f\u0928\u094d\u0926\u0941 \u0939\u0948","breadcrumb":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-21-1-mathematics-solution-class-11-chapter-21-dirghavrutta\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-21-1-mathematics-solution-class-11-chapter-21-dirghavrutta\/"]}]},{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-21-1-mathematics-solution-class-11-chapter-21-dirghavrutta\/#primaryimage","url":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-22-1-scaled.jpg","contentUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-22-1-scaled.jpg","width":1600,"height":901,"caption":"KC Sinha: Exercise 21.1- Mathematics Solution Class 11 Chapter 21 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