The product of the perpendicular from any point on the hyperbola x2a2−y2b

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Question

The product of the perpendicular from any point on the hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

to its asymptotes, is equal to

Options:

A . aba+b

B . a2b2a2+b2

C . 2a2b2a2+b2

D . None of these

Answer: Option B
:
B
Let

(a s e c θ, b t a n θ)

be the any point on the hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

The equations of the asymptotes of the given hyperbola are

\frac{x}{a} - \frac{y}{b} = 0

and

\frac{x}{a} + \frac{y}{b} = 0

Now,

p_{1}

= length of the perpendicular from

(a s e c θ, b t a n θ)

on

\frac{x}{a} + \frac{y}{b} = 0

= \frac{s e c θ - t a n θ}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}}

and

p_{2}

= length of the perpendicular from

(a s e c θ, b t a n θ)

\frac{x}{a} + \frac{y}{b} = 0

= \frac{s e c θ - t a n θ}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}}

∴

p_{1} p_{2} = \frac{s e c^{2} θ - t a n^{2} θ}{\frac{1}{a^{2}} + \frac{1}{b^{2}}} = \frac{a^{2} b^{2}}{a^{2} + b^{2}}

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