{"id":543938,"date":"2021-09-30T11:03:13","date_gmt":"2021-09-30T11:03:13","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=543938"},"modified":"2021-10-01T10:41:30","modified_gmt":"2021-10-01T10:41:30","slug":"rd-sharma-solutions-for-class-11-maths-chapter-27-hyperbola","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-27-hyperbola\/","title":{"rendered":"RD Sharma Solutions for Class 11 Maths Chapter 27\u2013Hyperbola"},"content":{"rendered":"\n

Class 11: Maths Chapter 27 solutions. Complete Class 11 Maths Chapter 27 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 11 Maths Chapter 27\u2013Hyperbola<\/h2>\n\n\n\n

RD Sharma 11th Maths Chapter 27, Class 11 Maths Chapter 27 solutions<\/p>\n\n\n\n

EXERCISE 27.1 PAGE NO: 27.13<\/p>\n\n\n\n

1. The equation of the directrix of a hyperbola is x \u2013 y + 3 = 0. Its focus is (-1, 1) and eccentricity 3. Find the equation of the hyperbola.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equation of the directrix of a hyperbola => x \u2013 y + 3 = 0.<\/p>\n\n\n\n

Focus = (-1, 1) and<\/p>\n\n\n\n

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

Now, let us find the equation of the hyperbola<\/p>\n\n\n\n

Let \u2018M\u2019 be the point on directrix and P(x, y) be any point of the hyperbola.<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

e = PF\/PM<\/p>\n\n\n\n

PF = ePM [where, e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix]<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

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

[We know that (a \u2013 b)2<\/sup> = a2<\/sup> + b2<\/sup> + 2ab &(a + b + c)2<\/sup> = a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ac]<\/p>\n\n\n\n

So, 2{x2<\/sup> + 1 + 2x + y2<\/sup> + 1 \u2013 2y} = 9{x2<\/sup> + y2<\/sup>+ 9 + 6x \u2013 6y \u2013 2xy}<\/p>\n\n\n\n

2x2<\/sup> + 2 + 4x + 2y2<\/sup> + 2 \u2013 4y = 9x2<\/sup> + 9y2<\/sup>+ 81 + 54x \u2013 54y \u2013 18xy<\/p>\n\n\n\n

2x2<\/sup> + 4 + 4x + 2y2<\/sup>\u2013 4y \u2013 9x2<\/sup> \u2013 9y2<\/sup> \u2013 81 \u2013 54x + 54y + 18xy = 0<\/p>\n\n\n\n

\u2013 7x2<\/sup> \u2013 7y2<\/sup> \u2013 50x + 50y + 18xy \u2013 77 = 0<\/p>\n\n\n\n

7(x2<\/sup> + y2<\/sup>) \u2013 18xy + 50x \u2013 50y + 77 = 0<\/p>\n\n\n\n

\u2234The equation of hyperbola is 7(x2<\/sup> + y2<\/sup>) \u2013 18xy + 50x \u2013 50y + 77 = 0<\/p>\n\n\n\n

2. Find the equation of the hyperbola whose<\/strong><\/p>\n\n\n\n

(i) focus is (0, 3), directrix is x + y \u2013 1 = 0 and eccentricity = 2<\/strong><\/p>\n\n\n\n

(ii) focus is (1, 1), directrix is 3x + 4y + 8 = 0 and eccentricity = 2<\/strong><\/p>\n\n\n\n

(iii) focus is (1, 1) directrix is 2x + y = 1 and eccentricity =\u221a3<\/strong><\/p>\n\n\n\n

(iv) focus is (2, -1), directrix is 2x + 3y = 1 and eccentricity = 2<\/strong><\/p>\n\n\n\n

(v) focus is (a, 0), directrix is 2x + 3y = 1 and eccentricity = 2<\/strong><\/p>\n\n\n\n

(vi)focus is (2, 2), directrix is x + y = 9 and eccentricity = 2<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>focus is (0, 3), directrix is x + y \u2013 1 = 0 and eccentricity = 2<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Focus = (0, 3)<\/p>\n\n\n\n

Directrix => x + y \u2013 1 = 0<\/p>\n\n\n\n

Eccentricity = 2<\/p>\n\n\n\n

Now, let us find the equation of the hyperbola<\/p>\n\n\n\n

Let \u2018M\u2019 be the point on directrix and P(x, y) be any point of the hyperbola.<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

e = PF\/PM<\/p>\n\n\n\n

PF = ePM [where, e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix]<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

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

[We know that (a \u2013 b)2<\/sup> = a2<\/sup> + b2<\/sup> + 2ab &(a + b + c)2<\/sup> = a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ac]<\/p>\n\n\n\n

So, 2{x2<\/sup> + y2<\/sup> + 9 \u2013 6y} = 4{x2<\/sup> + y2<\/sup> + 1 \u2013 2x \u2013 2y + 2xy}<\/p>\n\n\n\n

2x2<\/sup> + 2y2<\/sup> + 18 \u2013 12y = 4x2<\/sup> + 4y2<\/sup>+ 4 \u2013 8x \u2013 8y + 8xy<\/p>\n\n\n\n

2x2<\/sup> + 2y2<\/sup> + 18 \u2013 12y \u2013 4x2<\/sup> \u2013 4y2<\/sup> \u2013 4 \u2013 8x + 8y \u2013 8xy = 0<\/p>\n\n\n\n

\u2013 2x2<\/sup> \u2013 2y2<\/sup> \u2013 8x \u2013 4y \u2013 8xy + 14 = 0<\/p>\n\n\n\n

-2(x2<\/sup> + y2<\/sup> \u2013 4x + 2y + 4xy \u2013 7) = 0<\/p>\n\n\n\n

x2<\/sup> + y2<\/sup> \u2013 4x + 2y + 4xy \u2013 7 = 0<\/p>\n\n\n\n

\u2234The equation of hyperbola is x2<\/sup> + y2<\/sup> \u2013 4x + 2y + 4xy \u2013 7 = 0<\/p>\n\n\n\n

(ii) <\/strong>focus is (1, 1), directrix is 3x + 4y + 8 = 0 and eccentricity = 2<\/p>\n\n\n\n

Focus = (1, 1)<\/p>\n\n\n\n

Directrix => 3x + 4y + 8 = 0<\/p>\n\n\n\n

Eccentricity = 2<\/p>\n\n\n\n

Now, let us find the equation of the hyperbola<\/p>\n\n\n\n

Let \u2018M\u2019 be the point on directrix and P(x, y) be any point of the hyperbola.<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

e = PF\/PM<\/p>\n\n\n\n

PF = ePM [where, e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix]<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

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

[We know that (a \u2013 b)2<\/sup> = a2<\/sup> + b2<\/sup> + 2ab &(a + b + c)2<\/sup> = a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ac]<\/p>\n\n\n\n

25{x2<\/sup> + 1 \u2013 2x + y2<\/sup> + 1 \u2013 2y} = 4{9x2<\/sup> + 16y2<\/sup>+ 64 + 24xy + 64y + 48x}<\/p>\n\n\n\n

25x2<\/sup> + 25 \u2013 50x + 25y2<\/sup> + 25 \u2013 50y = 36x2<\/sup> + 64y2<\/sup> + 256 + 96xy + 256y + 192x<\/p>\n\n\n\n

25x2<\/sup> + 25 \u2013 50x + 25y2<\/sup> + 25 \u2013 50y \u2013 36x2<\/sup> \u2013 64y2<\/sup> \u2013 256 \u2013 96xy \u2013 256y \u2013 192x = 0<\/p>\n\n\n\n

\u2013 11x2<\/sup> \u2013 39y2<\/sup> \u2013 242x \u2013 306y \u2013 96xy \u2013 206 = 0<\/p>\n\n\n\n

11x2<\/sup> + 96xy + 39y2<\/sup> + 242x + 306y + 206 = 0<\/p>\n\n\n\n

\u2234The equation of hyperbola is11x2<\/sup> + 96xy + 39y2<\/sup> + 242x + 306y + 206 = 0<\/p>\n\n\n\n

(iii) <\/strong>focus is (1, 1) directrix is 2x + y = 1 and eccentricity =\u221a3<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Focus = (1, 1)<\/p>\n\n\n\n

Directrix => 2x + y = 1<\/p>\n\n\n\n

Eccentricity =\u221a3<\/p>\n\n\n\n

Now, let us find the equation of the hyperbola<\/p>\n\n\n\n

Let \u2018M\u2019 be the point on directrix and P(x, y) be any point of the hyperbola.<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

e = PF\/PM<\/p>\n\n\n\n

PF = ePM [where, e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix]<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

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

[We know that (a \u2013 b)2<\/sup> = a2<\/sup> + b2<\/sup> + 2ab &(a + b + c)2<\/sup> = a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ac]<\/p>\n\n\n\n

5{x2<\/sup> + 1 \u2013 2x + y2<\/sup> + 1 \u2013 2y} = 3{4x2<\/sup> + y2<\/sup>+ 1 + 4xy \u2013 2y \u2013 4x}<\/p>\n\n\n\n

5x2<\/sup> + 5 \u2013 10x + 5y2<\/sup> + 5 \u2013 10y = 12x2<\/sup> + 3y2<\/sup> + 3 + 12xy \u2013 6y \u2013 12x<\/p>\n\n\n\n

5x2<\/sup> + 5 \u2013 10x + 5y2<\/sup> + 5 \u2013 10y \u2013 12x2<\/sup> \u2013 3y2<\/sup> \u2013 3 \u2013 12xy + 6y + 12x = 0<\/p>\n\n\n\n

\u2013 7x2<\/sup> + 2y2<\/sup> + 2x \u2013 4y \u2013 12xy + 7 = 0<\/p>\n\n\n\n

7x2<\/sup> + 12xy \u2013 2y2<\/sup> \u2013 2x + 4y\u2013 7 = 0<\/p>\n\n\n\n

\u2234The equation of hyperbola is7x2<\/sup> + 12xy \u2013 2y2<\/sup> \u2013 2x + 4y\u2013 7 = 0<\/p>\n\n\n\n

(iv) <\/strong>focus is (2, -1), directrix is 2x + 3y = 1 and eccentricity = 2<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Focus = (2, -1)<\/p>\n\n\n\n

Directrix => 2x + 3y = 1<\/p>\n\n\n\n

Eccentricity = 2<\/p>\n\n\n\n

Now, let us find the equation of the hyperbola<\/p>\n\n\n\n

Let \u2018M\u2019 be the point on directrix and P(x, y) be any point of the hyperbola.<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

e = PF\/PM<\/p>\n\n\n\n

PF = ePM [where, e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix]<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

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

[We know that (a \u2013 b)2<\/sup> = a2<\/sup> + b2<\/sup> + 2ab &(a + b + c)2<\/sup> = a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ac]<\/p>\n\n\n\n

13{x2<\/sup> + 4 \u2013 4x + y2<\/sup> + 1 + 2y} = 4{4x2<\/sup> + 9y2<\/sup> + 1 + 12xy \u2013 6y \u2013 4x}<\/p>\n\n\n\n

13x2<\/sup> + 52 \u2013 52x + 13y2<\/sup> + 13 + 26y = 16x2<\/sup> + 36y2<\/sup> + 4 + 48xy \u2013 24y \u2013 16x<\/p>\n\n\n\n

13x2<\/sup> + 52 \u2013 52x + 13y2<\/sup> + 13 + 26y \u2013 16x2<\/sup> \u2013 36y2<\/sup> \u2013 4 \u2013 48xy + 24y + 16x = 0<\/p>\n\n\n\n

\u2013 3x2<\/sup> \u2013 23y2<\/sup> \u2013 36x + 50y \u2013 48xy + 61 = 0<\/p>\n\n\n\n

3x2<\/sup> + 23y2<\/sup> + 48xy + 36x \u2013 50y\u2013 61 = 0<\/p>\n\n\n\n

\u2234The equation of hyperbola is3x2<\/sup> + 23y2<\/sup> + 48xy + 36x \u2013 50y\u2013 61 = 0<\/p>\n\n\n\n

(v) <\/strong>focus is (a, 0), directrix is 2x + 3y = 1 and eccentricity = 2<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Focus = (a, 0)<\/p>\n\n\n\n

Directrix => 2x + 3y = 1<\/p>\n\n\n\n

Eccentricity = 2<\/p>\n\n\n\n

Now, let us find the equation of the hyperbola<\/p>\n\n\n\n

Let \u2018M\u2019 be the point on directrix and P(x, y) be any point of the hyperbola.<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

e = PF\/PM<\/p>\n\n\n\n

PF = ePM [where, e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix]<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

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

[We know that (a \u2013 b)2<\/sup> = a2<\/sup> + b2<\/sup> + 2ab &(a + b + c)2<\/sup> = a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ac]<\/p>\n\n\n\n

45{x2<\/sup> + a2<\/sup> \u2013 2ax + y2<\/sup>} = 16{4x2<\/sup> + y2<\/sup> + a2<\/sup> \u2013 4xy \u2013 2ay + 4ax}<\/p>\n\n\n\n

45x2<\/sup> + 45a2<\/sup> \u2013 90ax + 45y2<\/sup> = 64x2<\/sup> + 16y2<\/sup> + 16a2<\/sup> \u2013 64xy \u2013 32ay + 64ax<\/p>\n\n\n\n

45x2<\/sup> + 45a2<\/sup> \u2013 90ax + 45y2<\/sup> \u2013 64x2<\/sup> \u2013 16y2<\/sup> \u2013 16a2<\/sup> + 64xy + 32ay \u2013 64ax = 0<\/p>\n\n\n\n

19x2<\/sup> \u2013 29y2<\/sup> + 154ax \u2013 32ay \u2013 64xy \u2013 29a2<\/sup> = 0<\/p>\n\n\n\n

\u2234The equation of hyperbola is19x2<\/sup> \u2013 29y2<\/sup> + 154ax \u2013 32ay \u2013 64xy \u2013 29a2<\/sup> = 0<\/p>\n\n\n\n

(vi) <\/strong>focus is (2, 2), directrix is x + y = 9 and eccentricity = 2<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Focus = (2, 2)<\/p>\n\n\n\n

Directrix => x + y = 9<\/p>\n\n\n\n

Eccentricity = 2<\/p>\n\n\n\n

Now, let us find the equation of the hyperbola<\/p>\n\n\n\n

Let \u2018M\u2019 be the point on directrix and P(x, y) be any point of the hyperbola.<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

e = PF\/PM<\/p>\n\n\n\n

PF = ePM [where, e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix]<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

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

[We know that (a \u2013 b)2<\/sup> = a2<\/sup> + b2<\/sup> + 2ab &(a + b + c)2<\/sup> = a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ac]<\/p>\n\n\n\n

x2<\/sup> + 4 \u2013 4x + y2<\/sup> + 4 \u2013 4y = 2{x2<\/sup> + y2<\/sup> + 81 + 2xy \u2013 18y \u2013 18x}<\/p>\n\n\n\n

x2<\/sup> \u2013 4x + y2<\/sup> + 8 \u2013 4y = 2x2<\/sup> + 2y2<\/sup> + 162 + 4xy \u2013 36y \u2013 36x<\/p>\n\n\n\n

x2<\/sup> \u2013 4x + y2<\/sup> + 8 \u2013 4y \u2013 2x2<\/sup> \u2013 2y2<\/sup> \u2013 162 \u2013 4xy + 36y + 36x = 0<\/p>\n\n\n\n

\u2013 x2<\/sup> \u2013 y2<\/sup> + 32x + 32y + 4xy \u2013 154 = 0<\/p>\n\n\n\n

x2<\/sup> + 4xy + y2<\/sup> \u2013 32x \u2013 32y + 154 = 0<\/p>\n\n\n\n

\u2234The equation of hyperbola isx2<\/sup> + 4xy + y2<\/sup> \u2013 32x \u2013 32y + 154 = 0<\/p>\n\n\n\n

3. Find the eccentricity, coordinates of the foci, equations of directrices and length of the latus-rectum of the hyperbola.
(i) 9x2<\/sup> \u2013 16y2<\/sup> = 144<\/strong><\/p>\n\n\n\n

(ii) 16x2<\/sup> \u2013 9y2<\/sup> = -144<\/strong><\/p>\n\n\n\n

(iii) 4x2<\/sup> \u2013 3y2<\/sup> = 36<\/strong><\/p>\n\n\n\n

(iv) 3x2<\/sup> \u2013 y2<\/sup> = 4<\/strong><\/p>\n\n\n\n

(v) 2x2<\/sup> \u2013 3y2<\/sup> = 5<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>9x2<\/sup> \u2013 16y2<\/sup> = 144<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equation => 9x2<\/sup> \u2013 16y2<\/sup> = 144<\/p>\n\n\n\n

The equation can be expressed as:<\/p>\n\n\n\n

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

The obtained equation is of the form<\/p>\n\n\n\n

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

Where, a2<\/sup> = 16, b2<\/sup> = 9 i.e., a = 4 and b = 3<\/p>\n\n\n\n

Eccentricity is given by:<\/p>\n\n\n\n

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

Foci: The coordinates of the foci are (0, \u00b1be)<\/p>\n\n\n\n

(0, \u00b1be) = (0, \u00b14(5\/4))<\/p>\n\n\n\n

= (0, \u00b15)<\/p>\n\n\n\n

The equation of directrices is given as:<\/p>\n\n\n\n

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

5x \u2213 16 = 0<\/p>\n\n\n\n

The length of latus-rectum is given as:<\/p>\n\n\n\n

2b2<\/sup>\/a<\/p>\n\n\n\n

= 2(9)\/4<\/p>\n\n\n\n

= 9\/2<\/p>\n\n\n\n

(ii) <\/strong>16x2<\/sup> \u2013 9y2<\/sup> = -144<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equation => 16x2<\/sup> \u2013 9y2<\/sup> = -144<\/p>\n\n\n\n

The equation can be expressed as:<\/p>\n\n\n\n

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

The obtained equation is of the form<\/p>\n\n\n\n

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

Where, a2<\/sup> = 9, b2<\/sup> = 16 i.e., a = 3 and b = 4<\/p>\n\n\n\n

Eccentricity is given by:<\/p>\n\n\n\n

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

Foci: The coordinates of the foci are (0, \u00b1be)<\/p>\n\n\n\n

(0, \u00b1be) = (0, \u00b14(5\/4))<\/p>\n\n\n\n

= (0, \u00b15)<\/p>\n\n\n\n

The equation of directrices is given as:<\/p>\n\n\n\n

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

The length of latus-rectum is given as:<\/p>\n\n\n\n

2a2<\/sup>\/b<\/p>\n\n\n\n

= 2(9)\/4<\/p>\n\n\n\n

= 9\/2<\/p>\n\n\n\n

(iii) <\/strong>4x2<\/sup> \u2013 3y2<\/sup> = 36<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equation => 4x2<\/sup> \u2013 3y2<\/sup> = 36<\/p>\n\n\n\n

The equation can be expressed as:<\/p>\n\n\n\n

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

The obtained equation is of the form<\/p>\n\n\n\n

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

Where, a2<\/sup> = 9, b2<\/sup> = 12 i.e., a = 3 and b = \u221a12<\/p>\n\n\n\n

Eccentricity is given by:<\/p>\n\n\n\n

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

Foci: The coordinates of the foci are (\u00b1ae, 0)<\/p>\n\n\n\n

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

(\u00b1ae, 0) = (\u00b1\u221a<\/strong>21, 0)<\/p>\n\n\n\n

The equation of directrices is given as:<\/p>\n\n\n\n

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

The length of latus-rectum is given as:<\/p>\n\n\n\n

2b2<\/sup>\/a<\/p>\n\n\n\n

= 2(12)\/3<\/p>\n\n\n\n

= 24\/3<\/p>\n\n\n\n

= 8<\/p>\n\n\n\n

(iv) <\/strong>3x2<\/sup> \u2013 y2<\/sup> = 4<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equation => 3x2<\/sup> \u2013 y2<\/sup> = 4<\/p>\n\n\n\n

The equation can be expressed as:<\/p>\n\n\n\n

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

The obtained equation is of the form<\/p>\n\n\n\n

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

Where, a = 2\/\u221a3 and b = 2<\/p>\n\n\n\n

Eccentricity is given by:<\/p>\n\n\n\n

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

Foci: The coordinates of the foci are (\u00b1ae, 0)<\/p>\n\n\n\n

(\u00b1ae, 0) = \u00b1(2\/\u221a3)(2) = \u00b14\/\u221a3<\/p>\n\n\n\n

(\u00b1ae, 0) = (\u00b14\/\u221a3, 0)<\/p>\n\n\n\n

The equation of directrices is given as:<\/p>\n\n\n\n

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

The length of latus-rectum is given as:<\/p>\n\n\n\n

2b2<\/sup>\/a<\/p>\n\n\n\n

= 2(4)\/[2\/\u221a3]<\/p>\n\n\n\n

= 4\u221a3<\/p>\n\n\n\n

(v) <\/strong>2x2<\/sup> \u2013 3y2<\/sup> = 5<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equation => 2x2<\/sup> \u2013 3y2<\/sup> = 5<\/p>\n\n\n\n

The equation can be expressed as:<\/p>\n\n\n\n

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

The obtained equation is of the form<\/p>\n\n\n\n

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

Where, a = \u221a5\/\u221a2 and b = \u221a5\/\u221a3<\/p>\n\n\n\n

Eccentricity is given by:<\/p>\n\n\n\n

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

Foci: The coordinates of the foci are (\u00b1ae, 0)<\/p>\n\n\n\n

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

(\u00b1ae, 0) = (\u00b15\/\u221a6, 0)<\/p>\n\n\n\n

The equation of directrices is given as:<\/p>\n\n\n\n

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

The length of latus-rectum is given as:<\/p>\n\n\n\n

2b2<\/sup>\/a<\/p>\n\n\n\n

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

4. Find the axes, eccentricity, latus-rectum and the coordinates of the foci of the hyperbola 25x2<\/sup> \u2013 36y2<\/sup> = 225.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equation=> 25x2<\/sup> \u2013 36y2<\/sup> = 225<\/p>\n\n\n\n

The equation can be expressed as:<\/p>\n\n\n\n

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

The obtained equation is of the form<\/p>\n\n\n\n

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

Where, a = 3 and b = 5\/2<\/p>\n\n\n\n

Eccentricity is given by:<\/p>\n\n\n\n

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

Foci: The coordinates of the foci are (\u00b1ae, 0)<\/p>\n\n\n\n

(\u00b1ae, 0) = \u00b13 (\u221a61\/6) = \u00b1 \u221a61\/2<\/p>\n\n\n\n

(\u00b1ae, 0) = (\u00b1 \u221a61\/2, 0)<\/p>\n\n\n\n

The equation of directrices is given as:<\/p>\n\n\n\n

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

The length of latus-rectum is given as:<\/p>\n\n\n\n

2b2<\/sup>\/a<\/p>\n\n\n\n

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

\u2234 Transverse axis = 6, conjugate axis = 5, e = \u221a61\/6, LR = 25\/6, foci = (\u00b1 \u221a61\/2, 0)<\/p>\n\n\n\n

5. Find the centre, eccentricity, foci and directions of the hyperbola
(i) 16x2<\/sup> \u2013 9y2<\/sup> + 32x + 36y \u2013 164 = 0<\/strong><\/p>\n\n\n\n

(ii) x2<\/sup> \u2013 y2<\/sup> + 4x = 0<\/strong><\/p>\n\n\n\n

(iii) x2<\/sup> \u2013 3y2<\/sup> \u2013 2x = 8<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>16x2<\/sup> \u2013 9y2<\/sup> + 32x + 36y \u2013 164 = 0<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equation => 16x2<\/sup> \u2013 9y2<\/sup> + 32x + 36y \u2013 164 = 0<\/p>\n\n\n\n

Let us find the centre, eccentricity, foci and directions of the hyperbola<\/p>\n\n\n\n

By using the given equation<\/p>\n\n\n\n

16x2<\/sup> \u2013 9y2<\/sup> + 32x + 36y \u2013 164 = 0<\/p>\n\n\n\n

16x2<\/sup> + 32x + 16 \u2013 9y2<\/sup> + 36y \u2013 36 \u2013 16 + 36 \u2013 164 = 0<\/p>\n\n\n\n

16(x2<\/sup> + 2x + 1) \u2013 9(y2<\/sup> \u2013 4y + 4) \u2013 16 + 36 \u2013 164 = 0<\/p>\n\n\n\n

16(x2<\/sup> + 2x + 1) \u2013 9(y2<\/sup> \u2013 4y + 4) \u2013 144 = 0<\/p>\n\n\n\n

16(x + 1)2<\/sup> \u2013 9(y \u2013 2)2<\/sup> = 144<\/p>\n\n\n\n

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

Here, center of the hyperbola is (-1, 2)<\/p>\n\n\n\n

So, let x + 1 = X and y \u2013 2 = Y<\/p>\n\n\n\n

The obtained equation is of the form<\/p>\n\n\n\n

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

Where, a = 3 and b = 4<\/p>\n\n\n\n

Eccentricity is given by:<\/p>\n\n\n\n

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

Foci: The coordinates of the foci are (\u00b1ae, 0)<\/p>\n\n\n\n

X = \u00b15 and Y = 0<\/p>\n\n\n\n

x + 1 = \u00b15 and y \u2013 2 = 0<\/p>\n\n\n\n

x = \u00b15 \u2013 1 and y = 2<\/p>\n\n\n\n

x = 4, -6 and y = 2<\/p>\n\n\n\n

So, Foci: (4, 2) (-6, 2)<\/p>\n\n\n\n

Equation of directrix are:<\/p>\n\n\n\n

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

\u2234 The center is (-1, 2), eccentricity (e) = 5\/3, Foci = (4, 2) (-6, 2), Equation of directrix = 5x \u2013 4 = 0 and 5x + 14 = 0<\/p>\n\n\n\n

(ii) <\/strong>x2<\/sup> \u2013 y2<\/sup> + 4x = 0<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equation => x2<\/sup> \u2013 y2<\/sup> + 4x = 0<\/p>\n\n\n\n

Let us find the centre, eccentricity, foci and directions of the hyperbola<\/p>\n\n\n\n

By using the given equation<\/p>\n\n\n\n

x2<\/sup> \u2013 y2<\/sup> + 4x = 0<\/p>\n\n\n\n

x2<\/sup> + 4x + 4 \u2013 y2<\/sup> \u2013 4 = 0<\/p>\n\n\n\n

(x + 2)2<\/sup> \u2013 y2<\/sup> = 4<\/p>\n\n\n\n

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

Here, center of the hyperbola is (2, 0)<\/p>\n\n\n\n

So, let x \u2013 2 = X<\/p>\n\n\n\n

The obtained equation is of the form<\/p>\n\n\n\n

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

Where, a = 2 and b = 2<\/p>\n\n\n\n

Eccentricity is given by:<\/p>\n\n\n\n

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

Foci: The coordinates of the foci are (\u00b1ae, 0)<\/p>\n\n\n\n

X = \u00b1 <\/strong>2\u221a2 and Y = 0<\/p>\n\n\n\n

X + 2 = \u00b1 <\/strong>2\u221a2 and Y = 0<\/p>\n\n\n\n

X= \u00b1 <\/strong>2\u221a2 \u2013 2 and Y = 0<\/p>\n\n\n\n

So, Foci = (\u00b1 <\/strong>2\u221a2 \u2013 2, 0)<\/p>\n\n\n\n

Equation of directrix are:<\/p>\n\n\n\n

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

\u2234 The center is (-2, 0), eccentricity (e) = \u221a2, Foci = (-2\u00b1 <\/strong>2\u221a2, 0), Equation of directrix =<\/p>\n\n\n\n

x + 2 = \u00b1<\/strong>\u221a2<\/p>\n\n\n\n

(iii) <\/strong>x2<\/sup> \u2013 3y2<\/sup> \u2013 2x = 8<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equation => x2<\/sup> \u2013 3y2<\/sup> \u2013 2x = 8<\/p>\n\n\n\n

Let us find the centre, eccentricity, foci and directions of the hyperbola<\/p>\n\n\n\n

By using the given equation<\/p>\n\n\n\n

x2<\/sup> \u2013 3y2<\/sup> \u2013 2x = 8<\/p>\n\n\n\n

x2<\/sup> \u2013 2x + 1 \u2013 3y2<\/sup> \u2013 1 = 8<\/p>\n\n\n\n

(x \u2013 1)2<\/sup> \u2013 3y2<\/sup> = 9<\/p>\n\n\n\n

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

Here, center of the hyperbola is (1, 0)<\/p>\n\n\n\n

So, let x \u2013 1 = X<\/p>\n\n\n\n

The obtained equation is of the form<\/p>\n\n\n\n

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

Where, a = 3 and b = \u221a3<\/p>\n\n\n\n

Eccentricity is given by:<\/p>\n\n\n\n

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

Foci: The coordinates of the foci are (\u00b1ae, 0)<\/p>\n\n\n\n

X = \u00b1 <\/strong>2\u221a3 and Y = 0<\/p>\n\n\n\n

X \u2013 1 = \u00b1 <\/strong>2\u221a3 and Y = 0<\/p>\n\n\n\n

X= \u00b1 <\/strong>2\u221a3 + 1 and Y = 0<\/p>\n\n\n\n

So, Foci = (1 \u00b1 <\/strong>2\u221a3, 0)<\/p>\n\n\n\n

Equation of directrix are:<\/p>\n\n\n\n

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

\u2234 The center is (1, 0), eccentricity (e) = 2\u221a3\/3, Foci = (1 \u00b1 <\/strong>2\u221a3, 0), Equation of directrix =<\/p>\n\n\n\n

X = 1\u00b1<\/strong>9\/2\u221a3<\/p>\n\n\n\n

6. Find the equation of the hyperbola, referred to its principal axes as axes of coordinates, in the following cases:
(i) the distance between the foci = 16 and eccentricity = \u221a2<\/strong><\/p>\n\n\n\n

(ii) conjugate axis is 5 and the distance between foci = 13<\/strong><\/p>\n\n\n\n

(iii) conjugate axis is 7 and passes through the point (3, -2)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>the distance between the foci = 16 and eccentricity = \u221a2<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Distance between the foci = 16<\/p>\n\n\n\n

Eccentricity = \u221a2<\/p>\n\n\n\n

Let us compare with the equation of the form<\/p>\n\n\n\n

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

Distance between the foci is 2ae and b2<\/sup> = a2<\/sup>(e2<\/sup> \u2013 1)<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

2ae = 16<\/p>\n\n\n\n

ae = 16\/2<\/p>\n\n\n\n

a\u221a2 = 8<\/p>\n\n\n\n

a = 8\/\u221a2<\/p>\n\n\n\n

a2<\/sup> = 64\/2<\/p>\n\n\n\n

= 32<\/p>\n\n\n\n

We know that, b2<\/sup> = a2<\/sup>(e2<\/sup> \u2013 1)<\/p>\n\n\n\n

So, b2<\/sup> = 32 [(\u221a2)2<\/sup> \u2013 1]<\/p>\n\n\n\n

= 32 (2 \u2013 1)<\/p>\n\n\n\n

= 32<\/p>\n\n\n\n

The Equation of hyperbola is given as<\/p>\n\n\n\n

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

x2<\/sup> \u2013 y2<\/sup> = 32<\/p>\n\n\n\n

\u2234 The Equation of hyperbola is x2<\/sup> \u2013 y2<\/sup> = 32<\/p>\n\n\n\n

(ii) <\/strong>conjugate axis is 5 and the distance between foci = 13<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Conjugate axis = 5<\/p>\n\n\n\n

Distance between foci = 13<\/p>\n\n\n\n

Let us compare with the equation of the form<\/p>\n\n\n\n

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

Distance between the foci is 2ae and b2<\/sup> = a2<\/sup>(e2<\/sup> \u2013 1)<\/p>\n\n\n\n

Length of conjugate axis is 2b<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

2b = 5<\/p>\n\n\n\n

b = 5\/2<\/p>\n\n\n\n

b2<\/sup> = 25\/4<\/p>\n\n\n\n

We know that, 2ae = 13<\/p>\n\n\n\n

ae = 13\/2<\/p>\n\n\n\n

a2<\/sup>e2<\/sup> = 169\/4<\/p>\n\n\n\n

b2<\/sup> = a2<\/sup>(e2<\/sup> \u2013 1)<\/p>\n\n\n\n

b2<\/sup> = a2<\/sup>e2<\/sup> \u2013 a2<\/sup><\/p>\n\n\n\n

25\/4 = 169\/4 \u2013 a2<\/sup><\/p>\n\n\n\n

a2<\/sup> = 169\/4 \u2013 25\/4<\/p>\n\n\n\n

= 144\/4<\/p>\n\n\n\n

= 36<\/p>\n\n\n\n

The Equation of hyperbola is given as<\/p>\n\n\n\n

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

\u2234 The Equation of hyperbola is 25x2<\/sup> \u2013 144y2<\/sup> = 900<\/p>\n\n\n\n

(iii)<\/strong> conjugate axis is 7 and passes through the point (3, -2)<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Conjugate axis = 7<\/p>\n\n\n\n

Passes through the point (3, -2)<\/p>\n\n\n\n

Conjugate axis is 2b<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

2b = 7<\/p>\n\n\n\n

b = 7\/2<\/p>\n\n\n\n

b2<\/sup> = 49\/4<\/p>\n\n\n\n

The Equation of hyperbola is given as<\/p>\n\n\n\n

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

Since it passes through points (3, -2)<\/p>\n\n\n\n

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

a2<\/sup> = 441\/65<\/p>\n\n\n\n

The equation of hyperbola is given as:<\/p>\n\n\n\n

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

\u2234 The Equation of hyperbola is 65x2<\/sup> \u2013 36y2<\/sup> = 441<\/p>\n\n\n\n

RD Sharma Solutions for Class 11 Maths Chapter 27: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 11 Maths Chapter 27\u2013Hyperbola<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 11 Maths Chapter 27\u2013Hyperbola PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 11 Maths :<\/strong><\/h2>\n\n\n\n