Question

Which one of the following points does not lie on the normal to the hyperbola, $\frac{x^2}{16}-\frac{y^2}{9}=1$ drawn at the point $(8,3\sqrt{3})$?

• A. $(13,-\frac{1}{\sqrt{3}})$
• B. $(12,\frac{1}{\sqrt{3}})$
• C. $(11,\sqrt{3})$
• D. $(10,\sqrt{3})$

Answer 1

Answer: (D)

$(10,\sqrt{3})$

$\frac{x^2}{16}-\frac{y^2}{9}=1$ $\Rightarrow \frac{2x}{16}-\frac{2y}{9}\frac{dy}{dx}=0$

At $(8,3\sqrt{3})$

$\frac{2\times 8}{16}-\frac{2\times 3\sqrt{3}}{9}\frac{dy}{dx}=0\Rightarrow \frac{3}{2\sqrt{3}}=\frac{dy}{dx}$

Therefore slope of normal $=-\frac{1}{\frac{dy}{dx}}$

Equation of normal at $(8,3\sqrt{3})$,

$y-3\sqrt{3}=-\frac{2}{\sqrt{3}}(x-8)$

Clearly, option (D) does not lies on it.