Q. The equation of the normal to the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ at $( - 4,\ 0)$ is
A) $2x - 3y = 1$
B) $x = 0$
C) $x = 1$
D) $y = 0$
Explanation
(d)
Here, $a^{2} = 16,\ b^{2} = 9$
The equation of normal at the point $( - 4,\ 0)$ is
$\frac{16x}{- 4} + \frac{9y}{0} = 16 + 9\ \left\lbrack \because\frac{a^{2}x}{x_{1}} + \frac{b^{2}y}{y_{1}} = a^{2} + b^{2} \right\rbrack$
$\Rightarrow \ \ \frac{9y}{0} = 25 + \frac{16x}{4}\ \Rightarrow \ y = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.