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)

Andhra Pradesh Board Solutions for Exercise 1: Exercise 5(a)

Author:Andhra Pradesh Board

Andhra Pradesh Board Mathematics Solutions for Exercise - Andhra Pradesh 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 Andhra Pradesh Board Solutions for Exercise 1: Exercise 5(a) with Hints & Solutions

MEDIUM

12th Andhra Pradesh 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 Andhra Pradesh 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 Andhra Pradesh 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 Andhra Pradesh 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 Andhra Pradesh 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 Andhra Pradesh 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 Andhra Pradesh 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 Andhra Pradesh 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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