Show that the locus of feet of the perpendiculars drawn from foci to any tangent of the hyperbola x 2 a 2 − y 2 b 2 = 1 is the auxiliary circle of the hyperbola.

Equation of hyperbola is x 2 a 2 − y 2 b 2 = 1 . Equation of tangent to hyperbola is y = m x ± a 2 m 2 − b 2 . ⇒ y − m x = ± a 2 m 2 − b 2 . . . 1 Slopes of the ⊥ line = - 1 m Equation of the line ⊥ to 1 and passing through foci ( ± a e , 0 ) is y = − 1 m ( x ± a e ) ⇒ m y = − ( x ± a e ) ⇒ x + m y = ± a e . . . 2 Squaring and adding 1 and 2 y 2 − m x 2 + ( x + m y ) 2 = a 2 m 2 − b 2 + a 2 e 2 ⇒ y 2 + m 2 x 2 − 2 m x y + x 2 + m 2 y 2 + 2 m x y = a 2 m 2 − a 2 e 2 − 1 + a 2 e 2 ∵ b 2 = a 2 e 2 − 1 ⇒ y 2 + m 2 x 2 + x 2 + m 2 y 2 = a 2 m 2 − a 2 e 2 + a 2 + a 2 e 2 ⇒ y 2 1 + m 2 + x 2 1 + m 2 = a 2 m 2 − a 2 e 2 + a 2 + a 2 e 2 ⇒ x 2 + y 2 1 + m 2 = a 2 1 + m 2 ⇒ x 2 + y 2 = a 2 which is the auxiliary circle.

Show that the angle between the two asymptotes of a hyperbola x 2 a 2 − y 2 b 2 = 1 is 2 tan − 1 ⁡ b a or 2 sec − 1 ⁡ ( e ) .

Let the equation of the hyperbola be x 2 a 2 − y 2 b 2 = 1 . The asymptotes of hyperbola are y = ± b a x where m 1 = b a and m 2 = − b a . If θ is the angle between asymptotes of hyperbola then tan ⁡ θ = m 1 − m 2 1 + m 1 m 2 = b a + b a 1 − b 2 a 2 = 2 a b a 2 a 2 a 2 − b 2 = 2 a b a 2 − b 2 Now, sec 2 ⁡ θ = 1 + tan 2 ⁡ θ = 1 + 4 a 2 b 2 a 2 − b 2 2 = a 2 − b 2 2 + 4 a 2 b 2 a 2 − b 2 2 = a 4 + b 4 - 2 a 2 b 2 + 4 a 2 b 2 a 2 − b 2 2 = a 4 + b 4 + 2 a 2 b 2 a 2 − b 2 2 = a 2 + b 2 2 a 2 − b 2 2 ⇒ sec ⁡ θ = a 2 + b 2 a 2 − b 2 ⇒ cos ⁡ θ = a 2 − b 2 a 2 + b 2 ∵ sec A = 1 cos A Also, we have cos ⁡ θ 2 = 1 + cos ⁡ θ 2 = 1 + a 2 − b 2 a 2 + b 2 2 = a 2 + b 2 + a 2 - b 2 a 2 + b 2 2 = 2 a 2 a 2 + b 2 × 1 2 = a 2 a 2 + b 2 = a 2 a 2 + a 2 e 2 − 1 ∵ b 2 = a 2 e 2 - 1 = a 2 a 2 + a 2 e 2 − a 2 = a a e = 1 e ⇒ sec ⁡ θ 2 = e ⇒ θ = 2 sec − 1 ⁡ ( e ) Also, we have tan ⁡ θ 2 = 1 − cos ⁡ θ 1 + cos ⁡ θ = 1 − a 2 − b 2 a 2 + b 2 1 + a 2 − b 2 a 2 + b 2 = a 2 + b 2 − a 2 − b 2 a 2 + b 2 + a 2 − b 2 = 2 b 2 2 a 2 = b a ⇒ θ 2 = tan − 1 ⁡ b a ⇒ θ = 2 tan − 1 ⁡ b a Hence proved.

E M B I B E

Intermediate Second Year Mathematics Paper 2B>Coordinate Geometry>Chapter 5 - Hyperbola>Exercise 5(a)

Telangana Board Solutions for Exercise 1: Exercise 5(a)

Author:Telangana Board

Telangana Board Mathematics Solutions for Exercise - Telangana Board Solutions for Exercise 1: Exercise 5(a)

Attempt the practice questions from Exercise 1: Exercise 5(a) with hints and solutions to strengthen your understanding. Intermediate Second Year Mathematics Paper 2B solutions are prepared by Experienced Embibe Experts.

Questions from Telangana Board Solutions for Exercise 1: Exercise 5(a) with Hints & Solutions

MEDIUM

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2B>Chapter 5 - Hyperbola>Exercise 5(a)>Q I

If $3 x - 4 y + k = 0$ is a tangent to $x^{2} - 4 y^{2} = 5$ find the value of $k$ .

MEDIUM

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2B>Chapter 5 - Hyperbola>Exercise 5(a)>Q II

Find the equation of the tangent to the hyperbola $x^{2} - 4 y^{2} = 4$ which is parallel to $x + 2 y = 0$ .

MEDIUM

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2B>Chapter 5 - Hyperbola>Exercise 5(a)>Q II

Find the equation of the tangent to the hyperbola $x^{2} - 4 y^{2} = 4$ which is perpendicular to line $x + 2 y = 0$ .

HARD

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2B>Chapter 5 - Hyperbola>Exercise 5(a)>Q II

Find the equations of tangent drawn to hyperbola $2 x^{2} - 3 y^{2} = 6 through (- 2, 1)$ .

HARD

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2B>Chapter 5 - Hyperbola>Exercise 5(a)>Q II

Prove that the product of the perpendicular distances from any point on a hyperbola to its asymptotes is constant.

HARD

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2B>Chapter 5 - Hyperbola>Exercise 5(a)>Q III

Tangents to hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ make angles $θ_{1}, θ_{2}$ with transverse axis of a hyperbola. Show that the point of intersection of these tangents lies on the curve $2 x y = k (x^{2} - a^{2})$ when $\tan θ_{1} + \tan θ_{2} = k .$

HARD

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2B>Chapter 5 - Hyperbola>Exercise 5(a)>Q III

Show that the locus of feet of the perpendiculars drawn from foci to any tangent of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is the auxiliary circle of the hyperbola.

HARD

12th Telangana Board

IMPORTANT

Intermediate Second Year Mathematics Paper 2B>Chapter 5 - Hyperbola>Exercise 5(a)>Q III

Show that the angle between the two asymptotes of a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is $2 \tan^{- 1} (\frac{b}{a})$ or $2 \sec^{- 1} (e)$ .

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