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RD Sharma solutions for Class 11 Mathematics chapter 27 - Hyperbola

Chapter 27 : Hyperbola

Pages 13 - 14

Q 1 | Page 13

The equation of the directrix of a hyperbola is x − y + 3 = 0. Its focus is (−1, 1) and eccentricity 3. Find the equation of the hyperbola.

Q 2.1 | Page 13

Find the equation of the hyperbola whose focus is (0, 3), directrix is x + y − 1 = 0 and eccentricity = 2 .

Q 2.2 | Page 13

Find the equation of the hyperbola whose focus is (1, 1), directrix is 3x + 4y + 8 = 0 and eccentricity = 2 .

Q 2.3 | Page 13

Find the equation of the hyperbola whose focus is (1, 1) directrix is 2x + y = 1 and eccentricity = $\sqrt{3}$.

Q 2.4 | Page 13

Find the equation of the hyperbola whose focus is (2, −1), directrix is 2x + 3y = 1 and eccentricity = 2 .

Q 2.5 | Page 13

Find the equation of the hyperbola whose focus is (a, 0), directrix is 2x − y + a = 0 and eccentricity = $\frac{4}{3}$.

Q 2.6 | Page 13

Find the equation of the hyperbola whose focus is (2, 2), directrix is x + y = 9 and eccentricity = 2.

Q 3.1 | Page 13

Find the eccentricity, coordinates of the foci, equation of directrice and length of the latus-rectum of the hyperbola .

9x2 − 16y2 = 144

Q 3.2 | Page 13

Find the eccentricity, coordinates of the foci, equation of directrice and length of the latus-rectum of the hyperbola .

16x2 − 9y2 = −144

Q 3.3 | Page 13

Find the eccentricity, coordinates of the foci, equation of directrice and length of the latus-rectum of the hyperbola .

4x2 − 3y2 = 36

Q 3.4 | Page 13

Find the eccentricity, coordinates of the foci, equation of directrice and length of the latus-rectum of the hyperbola .

3x2 − y2 = 4

Q 3.5 | Page 13

Find the eccentricity, coordinates of the foci, equation of directrice and length of the latus-rectum of the hyperbola .

2x2 − 3y2 = 5.

Q 4 | Page 13

Find the axes, eccentricity, latus-rectum and the coordinates of the foci of the hyperbola 25x2 − 36y2 = 225.

Q 5.1 | Page 13

Find the centre, eccentricity, foci and directrice of the hyperbola .

16x2 − 9y2 + 32x + 36y − 164 = 0

Q 5.2 | Page 13

Find the centre, eccentricity, foci and directrice of the hyperbola .

x2 − y2 + 4x = 0

Q 5.3 | Page 13

Find the centre, eccentricity, foci and directrice of the hyperbola .

x2 − 3y2 − 2x = 8.

Q 6.1 | Page 13

Find the equation of the hyperbola, referred to its principal axes as axes of coordinates, in  the distance between the foci = 16 and eccentricity = $\sqrt{2}$.

Q 6.2 | Page 13

Find the equation of the hyperbola, referred to its principal axes as axes of coordinates, in  the  conjugate axis is 5 and the distance between foci = 13 .

Q 6.3 | Page 13

Find the equation of the hyperbola, referred to its principal axes as axes of coordinates, in  the conjugate axis is 7 and passes through the point (3, −2).

Q 7.1 | Page 14

Find the equation of the hyperbola whose foci are (6, 4) and (−4, 4) and eccentricity is 2.

Q 7.2 | Page 14

Find the equation of the hyperbola whose vertices are (−8, −1) and (16, −1) and focus is (17, −1).

Q 7.3 | Page 14

Find the equation of the hyperbola whose  foci are (4, 2) and (8, 2) and eccentricity is 2.

Q 7.4 | Page 14

Find the equation of the hyperbola whose vertices are at (0 ± 7) and foci at $\left( 0, \pm \frac{28}{3} \right)$ .

Q 7.5 | Page 14

Find the equation of the hyperbola whose vertices are at (± 6, 0) and one of the directrices is x = 4.

Q 7.6 | Page 14

Find the equation of the hyperbola whose foci at (± 2, 0) and eccentricity is 3/2.

Q 8 | Page 14

Find the eccentricity of the hyperbola, the length of whose conjugate axis is $\frac{3}{4}$ of the length of transverse axis.

Q 9.1 | Page 14

Find the equation of the hyperboala whose focus is at (5, 2), vertex at (4, 2) and centre at (3, 2).

Q 9.2 | Page 14

Find the equation of the hyperboala whose focus is at (4, 2), centre at (6, 2) and e = 2.

Q 10 | Page 14

If P is any point on the hyperbola whose axis are equal, prove that SP. S'P = CP2.

Q 11.01 | Page 14

Find the equation of the hyperbola satisfying the given condition :

vertices (± 2, 0), foci (± 3, 0)

Q 11.02 | Page 14

Find the equation of the hyperbola satisfying the given condition :

vertices (0, ± 5), foci (0, ± 8)

Q 11.03 | Page 14

Find the equation of the hyperbola satisfying the given condition :

vertices (0, ± 3), foci (0, ± 5)

Q 11.04 | Page 14

Find the equation of the hyperbola satisfying the given condition :

foci (± $3\sqrt{5}$ 0), the latus-rectum = 8

Q 11.05 | Page 14

Find the equation of the hyperbola satisfying the given condition :

foci (0, ± 13), conjugate axis = 24

Q 11.06 | Page 14

find the equation of the hyperbola satisfying the given condition:

foci (± $3\sqrt{5}$  0), the latus-rectum = 8

Q 11.07 | Page 14

(vii)  find the equation of the hyperbola satisfying the given condition:

foci (± 4, 0), the latus-rectum = 12

Q 11.08 | Page 14

find the equation of the hyperbola satisfying the given condition:

vertices (± 7, 0), $e = \frac{4}{3}$

Q 11.09 | Page 14

find the equation of the hyperbola satisfying the given condition:

foci (0, ± $\sqrt{10}$, passing through (2, 3).

Q 11.1 | Page 14

find the equation of the hyperbola satisfying the given condition:

foci (0, ± 12), latus-rectum = 36

Q 12 | Page 14

If the distance between the foci of a hyperbola is 16 and its ecentricity is $\sqrt{2}$,then obtain its equation.

Q 13 | Page 14

Show that the set of all points such that the difference of their distances from (4, 0) and (− 4,0) is always equal to 2 represents a hyperbola.

Page 18

Q 1 | Page 18

Write the eccentricity of the hyperbola 9x2 − 16y2 = 144.

Q 2 | Page 18

Write the eccentricity of the hyperbola whose latus-rectum is half of its transverse axis.

Q 3 | Page 18

Write the coordinates of the foci of the hyperbola 9x2 − 16y2 = 144.

Q 4 | Page 18

Write the equation of the hyperbola of eccentricity $\sqrt{2}$,  if it is known that the distance between its foci is 16.

Q 5 | Page 18

If the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ and the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$ coincide, write the value of b2.

Q 6 | Page 18

Write the length of the latus-rectum of the hyperbola 16x2 − 9y2 = 144.

Q 7 | Page 18

If the latus-rectum through one focus of a hyperbola subtends a right angle at the farther vertex, then write the eccentricity of the hyperbola.

Q 8 | Page 18

Write the distance between the directrices of the hyperbola x = 8 sec θ, y = 8 tan θ.

Q 9 | Page 18

Write the equation of the hyperbola whose vertices are (± 3, 0) and foci at (± 5, 0).

Q 10 | Page 18

If e1 and e2 are respectively the eccentricities of the ellipse $\frac{x^2}{18} + \frac{y^2}{4} = 1$

and the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ then write the value of 2 e12 + e22.

Pages 18 - 20

Q 1 | Page 18

Equation of the hyperbola whose vertices are (± 3, 0) and foci at (± 5, 0), is

16x2 − 9y2 = 144

9x2 − 16y2 = 144

25x2 − 9y= 225

9x2 − 25y2 = 81

Q 2 | Page 18

If e1 and e2 are respectively the eccentricities of the ellipse $\frac{x^2}{18} + \frac{y^2}{4} = 1$ and the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ , then the relation between e1 and e2 is

3 e12 + e22 = 2

e12 + 2 e22 = 3

e12 + e22 = 3

e12 + 3 e22 = 2

Q 3 | Page 19

The distance between the directrices of the hyperbola x = 8 sec θ, y = 8 tan θ, is

$8\sqrt{2}$

$16\sqrt{2}$

$4\sqrt{2}$

$6\sqrt{2}$

Q 4 | Page 19

The equation of the conic with focus at (1, 1) directrix along x − y + 1 = 0 and eccentricity $\sqrt{2}$ is

xy = 1

2xy + 4x − 4y − 1= 0

x2 − y2 = 1

2xy − 4x + 4y + 1 = 0

Q 5 | Page 19

The eccentricity of the conic 9x2 − 16y2 = 144 is

$\frac{5}{4}$

$\frac{4}{3}$

$\frac{4}{5}$

$\sqrt{7}$

Q 6 | Page 19

A point moves in a plane so that its distances PA and PB from two fixed points A and B in the plane satisfy the relation PA − PB = k (k ≠ 0), then the locus of P is

a hyperbola

a branch of the hyperbola

a parabola

an ellipse

Q 7 | Page 19

The eccentricity of the hyperbola whose latus-rectum is half of its transverse axis, is

$\frac{1}{\sqrt{2}}$

$\sqrt{\frac{2}{3}}$

$\sqrt{\frac{3}{2}}$

none of these.

Q 8 | Page 19

The eccentricity of the hyperbola x2 − 4y2 = 1 is

$\frac{\sqrt{3}}{2}$

$\frac{\sqrt{5}}{2}$

$\frac{2}{\sqrt{3}}$

$\frac{2}{\sqrt{5}}$

Q 9 | Page 19

The difference of the focal distances of any point on the hyperbola is equal to

length of the conjugate axis

eccentricity

length of the transverse axis

Latus-rectum

Q 10 | Page 19

The foci of the hyperbola 9x2 − 16y2 = 144 are

(± 4, 0)

(0, ± 4)

(± 5, 0)

(0, ± 5)

Q 11 | Page 19

The distance between the foci of a hyperbola is 16 and its eccentricity is $\sqrt{2}$, then equation of the hyperbola is

x2 + y2 = 32

x2 − y2 = 16

x2 + y2 = 16

x2 − y2 = 32

Q 12 | Page 19

If e1 is the eccentricity of the conic 9x2 + 4y2 = 36 and e2 is the eccentricity of the conic 9x2 − 4y2 = 36, then

e12 − e22 = 2

2 < e22 − e12 < 3

e22 − e12 = 2

e22 − e12 > 3

Q 13 | Page 19

If the eccentricity of the hyperbola x2 − y2 sec2α = 5 is $\sqrt{3}$  times the eccentricity of the ellipse x2 sec2 α + y2 = 25, then α =

$\frac{\pi}{6}$

$\frac{\pi}{4}$

$\frac{\pi}{3}$

$\frac{\pi}{2}$

Q 14 | Page 19

The equation of the hyperbola whose foci are (6, 4) and (−4, 4) and eccentricity 2, is

$\frac{(x - 1 )^2}{25/4} - \frac{(y - 4 )^2}{75/4} = 1$

$\frac{(x + 1 )^2}{25/4} - \frac{(y + 4 )^2}{75/4} = 1$

$\frac{(x - 1 )^2}{75/4} - \frac{(y - 4 )^2}{25/4} = 1$

none of these

Q 15 | Page 19

The length of the straight line x − 3y = 1 intercepted by the hyperbola x2 − 4y2 = 1 is

$\frac{6}{\sqrt{5}}$

$3\sqrt{\frac{2}{5}}$

$6\sqrt{\frac{2}{5}}$

none of these

Q 16 | Page 20

The latus-rectum of the hyperbola 16x2 − 9y2 = 144 is

16/3

32/3

8/3

4/3

Q 17 | Page 20

The foci of the hyperbola 2x2 − 3y2 = 5 are

$( \pm 5/\sqrt{6}, 0)$

(± 5/6, 0)

$( \pm \sqrt{5}/6, 0)$

none of these

Q 18 | Page 20

The eccentricity the hyperbola $x = \frac{a}{2}\left( t + \frac{1}{t} \right), y = \frac{a}{2}\left( t - \frac{1}{t} \right)$ is

$\sqrt{2}$

$\sqrt{3}$

$2\sqrt{3}$

$3\sqrt{2}$

Q 19 | Page 20

The equation of the hyperbola whose centre is (6, 2) one focus is (4, 2) and of eccentricity 2 is

3 (x − 6)2 − (y −2)2 = 3

(x − 6)2 − 3 (y − 2)2 = 1

(x − 6)2 − 2 (y −2)2 = 1

2 (x − 6)2 − (y − 2)2 = 1

Q 20 | Page 20

The locus of the point of intersection of the lines $\sqrt{3}x - y - 4\sqrt{3}\lambda = 0 \text { and } \sqrt{3}\lambda + \lambda - 4\sqrt{3} = 0$  is a hyperbola of eccentricity

1

2

3

4

RD Sharma solutions for Class 11 Mathematics chapter 27 - Hyperbola

RD Sharma solutions for Class 11 Maths chapter 27 (Hyperbola) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Mathematics Class 11 solutions in a manner that help students grasp basic concepts better and faster.

Further, we at shaalaa.com are providing such solutions so that students can prepare for written exams. RD Sharma textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.

Concepts covered in Class 11 Mathematics chapter 27 Hyperbola are Standard Equation of a Circle, Latus Rectum, Standard Equation of Hyperbola, Eccentricity, Introduction of Hyperbola, Latus Rectum, Standard Equations of an Ellipse, Eccentricity, Special Cases of an Ellipse, Relationship Between Semi-major Axis, Semi-minor Axis and the Distance of the Focus from the Centre of the Ellipse, Introduction of Ellipse, Latus Rectum, Standard Equations of Parabola, Introduction of Parabola, Concept of Circle, Sections of a Cone.

Using RD Sharma Class 11 solutions Hyperbola exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in RD Sharma Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 11 prefer RD Sharma Textbook Solutions to score more in exam.

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