Answer
Intersection points: $P(2,2)$, $Q(-2, -2)$
Slope of the tangent to the curve $C$ at $P$ is $-1$
Slope of the tangent to the curve $C$ at $Q$ is $-1$
Tangents to the curve $C$ at point $P$ and point $Q$ are parallel.
Work Step by Step
Putting $y=x $ in the equation $ x^2-xy+y^2=4$
$x^2-x^2+x^2=4$
$x^2=4$ Imply that $x=2,-2$
Putting $x=2$ in the equation $ x^2-xy+y^2=4$
$4-2y+y^2=4$ Imply that $y^2-2y=0$ Imply that $y=0, -2$
Putting $x=-2$ in the equation $ x^2-xy+y^2=4$
$(-2) ^2+2y+y^2=4$ Imply that $4+2y+y^2=4$ Imply that $y^2+2y=0$
$y=0, -2$
Since $y=x$, possible values of points $P$ and $Q$ are $(2,2)$ and $(-2, -2)$
Differentiating $ x^2-xy+y^2=4$ with respect to x
$\frac{d(x^2-xy+y^2)} {dx}=\frac{d(4)}{dx}$
$\frac{d(x^2)} {dx}-\frac{d(xy)} {dx}+\frac{d(y^2)} {dx}=0$
$2x-[x\frac{dy}{dx}+y\frac{dx}{dx}]+\frac{d(y^2)}{dy} \frac{dy}{dx}=0$
$2x-x\frac{dy}{dx}-y+2y \frac{dy}{dx}=0$
$(-x+2y) \frac{dy}{dx}=y-2x$
$(2y-x) \frac{dy}{dx}=y-2x$ Imply that
$\frac{dy}{dx}=\frac{y-2x} {2y-x} $ .......... eq (1)
Putting $(x,y)= (2,2) $ in equation (1)
Slope of tangent at point $P$ $=\frac{dy}{dx}=\frac {2-4} {4-2} =-1$
Putting $(x,y)= (-2, -2) $ in equation (1)
Slope of tangent at point $Q$ $=\frac{dy}{dx}=\frac { -2+4} {-4+2} =-1$
Tangents to the curve $C$ at points $P$ and $Q$ are parallel.
