Question

Find the equation of the normals to the curve $2x^2-y^2=14$ which are parallel to the line $x+3y=6$.

Answer 1

Slope of the normal:$\frac{-1}{3}$

Now, $y^2=2x^2-14$

$2y\frac{dy}{dx}=4x$

$\frac{dy}{dx}=\frac{2x}{y}$

Hence slope of the nowmal: $\frac{-y}{2x}=\frac{-1}{3}$

$y=\frac{2x}{3}$

Putting this in the equation:

$2x^2-{\frac{4x}{9}}^2=14$

Solving: $x^2=9$

$x=+3,-3$

By putting this in the quation, $y=+2,-2$

Hence the equation of the normals having slope=$\frac{-1}{3}$ are:

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

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

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

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