Show that at one of these points the tangent to the curve is parallel to the $x$ axis

Question

Show that at one of these points $(4,18)$,$(4,-2)$ the tangent line to the curve $y^2-4xy-x^2=20$ is parallel to the $x$ axis.
Any help would be appreciated

Answer 1

Derivate your curve like the following:
$\frac{d}{dx}(y^2-4xy-x^2)=2y(y')-2x-4x(y')-4y=\frac{dy}{dx}(20)=0$
$\implies y'=\frac{x+2y}{y+2x}$
So the tangent line in point $P=(x,y)$ is parallel to the x-axis $\iff y'=\frac{x+2y}{y+2x}=0 \iff x+2y=0 \iff x=-2y$ so the point $(4,-2)$ has the property.

Answer 2

Hint:

Calculate the slope by differentiating the curve at both of these points. Slope of any tangent parallel to x-axis should be zero.