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
$\frac{16 x}{-4}+\frac{9 y}{0}=16+9$
$\Rightarrow \,\,\,\,\, \frac{9 y}{0}=25+\frac{16 x}{4}$
$\Rightarrow \,\,\,\,\,\,9 y=0$
$\Rightarrow \,\,\,\,y=0$