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

The correct option is D $y=0$

$\frac{x^2}{16}-\frac{y^2}{9}=1$

Differentiating above equation, we get

$\frac{2x}{16}-\frac{2y}{9}.\frac{dy}{dx}=0$

$⟹\frac{dy}{dx}=\frac{9x}{16y}$

$⟹m={\left|\frac{dy}{dx}\right|}_{(-4,0)}=\frac{9(-4)}{16\left(0\right)}=\infty$ is the slope of tangent

Slope of the normal is $-\frac{1}{m}=0$

Slope of the normal is zero so it is parallel to $y$-axis.

Equation of normal is $y-0=0(x-(-4\left)\right)$

$\therefore y=0$ is the equation of normal.