Question

The equation to 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

Answer 1

Answer: (D)

We know that, the equation of normal at the
point $\left(x_1,y_1\right)$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ 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, $a^2=16,b^2=9$
$\therefore$ The equation of normal at the point $(-4,0)$ is
$\frac{16x}{-4}+\frac{9y}{0}=16+9$
$\Rightarrow \frac{9y}{0}=25+\frac{16x}{4}$
$\Rightarrow 9y=0$
$\Rightarrow y=0$