## Rajasthan Board RBSE Class 11 Maths Chapter 12 Conic Section Ex 12.7

Question 1.

Find the length of axis, focus eccentricity latus rectum and equation of directrix of the hyperbola

9x^{2} – 16y^{2} = 144.

Solution:

Given equation,

9x^{2} – 16y^{2} = 144.

Question 2.

Find the equations of hyperbola whose :

(i) Focus is (2, 1), directrix x + 2y – 1 = 0 and eccentricity is 2

(ii) Focus is (1,2), directrix 2x + y = 1 and eccentricity is \(\sqrt { 3 }\) .

Solution:

(i) Let (x, y) be any point at hyperbola then by definition, distance of point (x, y) from focus

= e × Distance of directrix from (x, y)

This is required equation

(ii) Let (x, y) be any point at hyperbola then by definition, distance of point (x, y) from focus

= e × Distance of point from (x, y) from focus

Question 3.

Find the vertex, focus, latus rectum and eccentricity of hyperbola

x^{2} – 6x – 4y^{2} – 16y – 11 = 0.

Solution:

Equation of hyperbola,

x^{2} – 6x – 4y^{2} – 16y – 11 = 0

⇒ (x^{2} – 2 × 3 × x + 3^{2} – 3^{2}) – 4(y^{2 }+ 4y + 2^{2} – 2^{2}) – 11 = 0

⇒ (x – 3)^{2} – 3^{2} – 4 (y + 2)^{2} + 16 – 11 = 0

⇒ (x – 3)^{2} – 4 (y + 2)^{2} + 16 – 11 – 9 = 0

Question 4.

Find the equation of hyperbola whose :

(i) Length of latus rectum is 8 and conjugate axis = 1/2 (Distance between focus)

(ii) Distance bewteen focus is 16 and conjugate axis is \(\sqrt { 2 }\)

(iii) Length of conjugate axis is 7 and passes through point (3, – 2).

Solution:

(i) Let equation of hyperbola,

(ii) Distance bewteen focus, 2ae = 16 conjugate axis 2b = \(\sqrt { 2 }\)

(iii) Conjugate axis 2b = 7

Question 5.

Prove that intersecting point of straight lines

\(\frac { x }{ a } \) – \(\frac { y }{ b } \) = m and \(\frac { x }{ a } \) + \(\frac { y }{ b } \) = \(\frac { 1 }{ m } \) is hyperbola.

a b a b m

Solution:

Given straight lines,

\(\frac { x }{ a } \) – \(\frac { y }{ b } \) = m ….(i)

and \(\frac { x }{ a } \) + \(\frac { y }{ b } \) = \(\frac { 1 }{ m } \) ….(ii)

Multiplying equation (i) and (ii)

(\(\frac { x }{ a } \) – \(\frac { y }{ b } \)) (\(\frac { x }{ a } \) + \(\frac { y }{ b } \)) = m × \(\frac { 1 }{ m } \)

This is equation of hyperbola.

Thus, locus of intersection point of two lines is hyperbola

Question 6.

Find the common point of hyperbola

5x^{2} – 9y^{2} = 45 and line y = x + 2.

Solution:

Equation of hyperbola,

5x^{2} – 9y^{2} = 45 …..(i)

and equation of line

y = x + 2 …..(ii)

From equation (i) and (ii),

5x^{2} – 9(x + 2)^{2} = 45

⇒ 5x^{2} – 9(x^{2} + 4x + 4) = 45

⇒ 5x^{2} – 9y^{2} – 36x – 36 – 45 = 0

⇒ -4x^{2} – 36x – 81 = 0

⇒ 4x^{2} + 36x + 81 = 0

On solving,

Question 7.

Prove that line lx + my = 1 will touch hyperbola

Solution:

Question 8.

Find the equation of tangents of hyperbola 4x^{2} – 9y^{2} = 1, which is parallel to line 4y = 5x + 7.

Solution:

Equation of hyperbola

4x^{2} – 9y^{2} = 1 ….(i)

Hence, required equation,

4y = 5x ⊥ \(\frac { 3 }{ 2 } \)

Question 9.

Prove that locus of foot of perpendicular drawn from focus at tangent of hyperbola is a circle.

Solution:

Let equation of hyperbola,

Eq. of line passing through (+c , 0) and perpendicular to line(ii)

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