Question

Find the equation of normal to the hyperbola $3x^2-y^2=1$ having slope $1/3$.

Answer 1

$3x^2-y^2=1$

Diff. wrt to $x$;

$6x-2y\frac{dy}{dx}=0$

$\therefore \frac{dy}{dx}=\frac{6x}{2y}=\frac{3x}{y}$

$\therefore \frac{dx}{dy}=\frac{-y}{3x}$

$\therefore \frac{-y}{3x}=\frac{1}{3};[x=-y]$

$\therefore 3(-y{)}^2-y^2=1$

$\therefore 2y^2=1$

$\therefore y=\frac{1}{\sqrt{2}}$

$\therefore x=\frac{-1}{\sqrt{2}}$

$\therefore$ Equation of Normal;

$y-\frac{1}{\sqrt{2}}=\frac{1}{3}(x+\frac{1}{\sqrt{2}})$

$\therefore 3y-\frac{3}{\sqrt{2}}=x+\frac{1}{\sqrt{2}}$

$\therefore 3y-x-2\sqrt{2}=0$