Question

A normal is drawn on the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ at a point given by parameter $\frac{\pi }{3}$.

What is the x-intercept of the normal.

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

For a hyperbola of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ the equation of normal is given by,

$x.acos\theta +y.bcot\theta =a^2+b^2$

here its better to keep this formula in mind. If not,

you can always gp for the long way of getting the normal by taking slope and point.

But for copetitive exams it not advisable

In the given question,

$a=4;b=3,\theta =\frac{\pi }{3}$

$x.4.cos.\frac{\pi }{3}+y.3.cot.\frac{\pi }{3}=16+9$

$4.x.\frac{\sqrt{3}}{2}+3y\frac{1}{\sqrt{3}}=25$

When normal meets x-axis y=0

$i.e.,\left(x\right)2\sqrt{3}=25$

$\therefore x=\frac{25}{2\sqrt{3}}=$ required x intercept; hence option (d) is correct option