{"id":562917,"date":"2021-12-20T05:56:43","date_gmt":"2021-12-20T05:56:43","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=562917"},"modified":"2021-12-28T05:43:36","modified_gmt":"2021-12-28T05:43:36","slug":"maharashtra-board-solutions-class-11-arts-science-maths-part-1-chapter-7-conic-sections","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-11-arts-science-maths-part-1-chapter-7-conic-sections\/","title":{"rendered":"Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 1): Chapter 7- Conic Sections"},"content":{"rendered":"\n

Class 11: Maths Chapter 7 solutions. Complete Class 11 Maths Chapter 1 Notes.<\/p>\n\n\n\n

Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 1): Chapter 7- Conic Sections<\/h2>\n\n\n\n

Maharashtra Board 11th Maths Chapter 7, Class 11 Maths Chapter 7 solutions<\/p>\n\n\n\n

Ex 7.1<\/h4>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

ii) Given equation of the parabola is y2<\/sup> = -20x.
Comparing this equation with y2<\/sup> = -4ax, we get
\u21d2 4a = 20
\u21d2 a = 5
Co-ordinates of focus are S(-a, 0), i.e., S(-5, 0)
Equation of the directrix is x \u2013 a = 0
\u21d2 x \u2013 5 = 0
Length of latus rectum = 4a = 4(5) = 20
Co-ordinates of end points of latus rectum are (-a, 2a) and (-a, -2a),
\u21d2 (-5, 10) and (-5, -10).<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

\u2234 The required equation of the parabola is x2<\/sup> = -4(5)y, i.e., x2<\/sup> = -20y.<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Question 4.
Find the equation of the parabola whose vertex is O(0, 0) and focus at (-7, 0).
Solution:<\/strong>
Focus of the parabola is S(-7, 0) and vertex is O(0, 0).
Since focus lies on X-axis, it is the axis of the parabola.
Focus S(-7, 0) lies on the left-hand side of the origin.
It is a left-handed parabola.
Required parabola is y = -4ax.
Focus is S(-a, 0).
a = 7
\u2234 The required equation of the parabola is y2<\/sup> =-4(7)x, i.e., y2<\/sup> = -28x.<\/p>\n\n\n\n

Question 5.
Find the equation of the parabola with vertex at the origin, the axis along X-axis, and passing through the point
(i) (1, -6)
(ii) (2, 3)
Solution:<\/strong>
(i) Vertex of the parabola is at origin (0, 0) and its axis is along X-axis.
Equation of the parabola can be either y2<\/sup> = 4ax or y2<\/sup> = -4ax.
Since the parabola passes through (1, -6), it lies in the 4th quadrant.
Required parabola is y2<\/sup> = 4ax.
Substituting x = 1 and y = -6 in y2<\/sup> = 4ax, we get
\u21d2 (-6)2<\/sup> = 4a(1)
\u21d2 36 = 4a
\u21d2 a = 9
\u2234 The required equation of the parabola is y2<\/sup> = 4(9)x, i.e., y2<\/sup> = 36x.<\/p>\n\n\n\n

(ii) Vertex of the parabola is at origin (0, 0) and its axis is along X-axis.
Equation of the parabola can be either y2<\/sup> = 4ax or y2<\/sup> = -4ax.
Since the parabola passes through (2, 3), it lies in 1st quadrant.
\u2234 Required parabola is y2<\/sup> = 4ax.
Substituting x = 2 and y = 3 in y2<\/sup> = 4ax, we get
\u21d2 (3)2<\/sup> = 4a(2)
\u21d2 9 = 8a<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

\u21d2 a = 1
Focal distance of a point = x + a
Given, focal distance = 17
\u21d2 x + 1 = 17
\u21d2 x = 16
Substituting x = 16 in y2<\/sup> = 4x, we get
\u21d2 y2<\/sup> = 4(16)
\u21d2 y2<\/sup> = 64
\u21d2 y = \u00b18
\u2234 The co-ordinates of the point on the parabola are (16, 8) or (16, -8).<\/p>\n\n\n\n

Question 10.
Find the length of the latus rectum of the parabola y2<\/sup> = 4ax passing through the point (2, -6).
Solution:<\/strong>
Given equation of the parabola is y2<\/sup> = 4ax and it passes through point (2, -6).
Substituting x = 2 and y = -6 in y2<\/sup> = 4ax, we get
\u21d2 (-6)2<\/sup> = 4a(2)
\u21d2 4a = 18
\u2234 Length of latus rectum = 4a = 18 units<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Question 12.
If a parabolic reflector is 20 cm in diameter and 5 cm deep, find its focus.
Solution:<\/strong>
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Let LOM be the parabolic reflector such that LM is the diameter and ON is its depth.
It is given that ON = 5 cm and LM = 20 cm.
LN = 10 cm
Taking O as the origin, ON along X-axis and a line through O \u22a5 ON as Y-axis.
Let the equation of the reflector be y2<\/sup> = 4ax \u2026\u2026(i)
The point L has the co-ordinates (5, 10) and lies on parabola given by (i).
Substituting x = 5 and y = 10 in (i), we get
\u21d2 102<\/sup> = 4a(5)
\u21d2 100 = 20a
\u21d2 a = 5
Focus is at (a, 0), i.e., (5, 0)<\/p>\n\n\n\n

Question 13.
Find co-ordinates of focus, vertex, and equation of directrix and the axis of the parabola y = x2<\/sup> \u2013 2x + 3.
Solution:<\/strong>
Given equation of the parabola is y = x2<\/sup> \u2013 2x + 3
\u21d2 y = x2<\/sup> \u2013 2x + 1 + 2
\u21d2 y \u2013 2 = (x \u2013 1)2<\/sup>
\u21d2 (x \u2013 1)2<\/sup> = y \u2013 2
Comparing this equation with X2<\/sup> = 4bY, we get
X = x \u2013 1, Y = y \u2013 2
\u21d2 4b = 1<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

Ex 7.2<\/h4>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

Length of minor axis = 2b = 2\u221a3
Lengths of the principal axes are 4 and 2\u221a3.<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Since a > b,
X-axis is the major axis and Y-axis is the minor axis.<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

Ex 7.3<\/h4>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

\u21d2 b = 5 and a = 3
Length of transverse axis = 2b = 2(5) = 10
Length of conjugate axis = 2a = 2(3) = 6
Co-ordinates of vertices are B(0, b) and B\u2019 (0, -b),
i.e., B(0, 5) and B\u2019 (0, -5)
We know that<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

Solution:
<\/strong>Let e be the eccentricity of a hyperbola
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

Question 5.
Find the equation of the hyperbola referred to its principal axes:<\/strong><\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 1): Chapter 7- Conic Sections<\/p>\n\n\n\n

Download PDF<\/strong>: Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 1): Chapter 7- Conic Sections PDF<\/a><\/p>\n\n\n\n

Chapterwise Maharashtra Board Solutions Class 11 Arts & Science Maths (Part 1) :<\/strong><\/h2>\n\n\n\n