Q72 / 180 · KCET 2008 · Mathematics
The equation of the normal to the hyperbola $\frac{x^{2}}{16}-
\frac{y^{2}}{9}=1$ at $(-4,0)$ is
- A$2 \mathrm{x}-3 \mathrm{y}=1$
- B$\mathrm{x}=0$
- C$x=1$
- Correct answerD$y=0$
Solution
We know that, the equation of normal at the point $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ to the hyperbola $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$ is
$$
\frac{a^{2} x}{x_{1}}+\frac{b^{2} y}{y_{1}}=a^{2}+b^{2}
$$
Given equation is $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$
Here, $\quad a^{2}=16, b^{2}=9$
$\therefore$ The equation of normal at the point $(-4,0)$ is
$$
\begin{aligned}
\frac{16 x}{-4}+\frac{9 y}{0} &=16+9 \\
\Rightarrow \quad \frac{9 y}{0} &=25+\frac{16 x}{4} \\
9 y &=0 \Rightarrow y=0
\end{aligned}
$$
Hence, the correct answer is option: 'D'.