Question

If the line $x-y+2=0$ is a normal to the parabola $y^2-6y-4x+k=0$, then find $k$.

Answer 1

We have,

$x-y+2=0\left[slope=1\right]y^2-6y-4x+k=02y\frac{dy}{dx}-\frac{6dy}{dx}-4=0\frac{dy}{dx}=\frac{4}{2y-6}-\frac{dx}{dy}=\frac{6-2y}{4}=1y=1thatgivesx=-1fromwhichk=1$

Hence, this is the answer.