First slide

Hyperbola in conic sections

Question

The product of the perpendiculars from the foci on any tangent to the hyperbola $\frac{x^{2}}{64} - \frac{y^{2}}{9} = 1$ is

Moderate

A

8
B

9
C

16
D

18.

Solution

Equation of a tangent at $(8 \sec θ, 3 \tan θ)$ the hyperbola is $\frac{x}{8} \sec θ - \frac{y}{3} \tan θ = 1$
If $e$ is the eccentricity of the hyperbola; product of the perpendiculars
$\begin{array}{l} = \frac{(e \sec θ - 1) (- e \sec θ - 1)}{\frac{\sec^{2} θ}{64} + \frac{\tan^{2} θ}{9}} \\ = \frac{(1 - e^{2} \sec^{2} θ) \times (64 \times 9)}{9 \sec^{2} θ + 64 \tan^{2} θ} \\ = \frac{[1 - \frac{64 - 9}{64} \sec^{2} θ] (64 \times 9)}{9 \sec^{2} θ + 64 \tan^{2} θ} \\ = \frac{64 \tan^{2} θ + 9 \sec^{2} θ}{9 \sec^{2} θ + 64 \tan^{2} θ} \times 9 = 9 . \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App