Answer
$P (a, b)$ and $Q (−a, −b)$ both satisfy the equation $ x^2 − xy + y^2 = 4$, so they belong to the graph.
Slope of tangents to the curve at $P(a,b)=\frac{b-2a} {2b-a} $
Slope of tangent to the curve at $ Q (-a, -b) =\frac{b-2a} {2b-a} $
Tangents are parallel at points $P$ and $Q$
Work Step by Step
Equation of the curve C
$x^2-xy+y^2=4$ ............ eq (1)
Putting P(a,b) in equation (1)
$a^2-ab+b^2=4$ ......... eq (2)
Putting Q (-a, -b) in equation (1)
$ (-a) ^2-(-a) (-b) +(-b) ^2=4$
Or $ a^2+ab +b^2=4$ ........ eq (3)
Since eq (2) is true, eq (3) being the same equation is also true. Hence point $Q (-a, -b)$ also lies on the same curve $C$.
Also $\frac{dy}{dx}=\frac{y-2x} {2y-x} $ ............. eq (3)
Putting $P(a,b)$ in equation (3)
$\frac{dy}{dx}=\frac{b-2a} {2b-a} $
Putting $P (-a, -b)$ in equation (3)
$\frac{dy}{dx}=\frac{-b+2a} {-2b+a} =\frac {-(b-2a)} {-(2b-a)} =\frac{b-2a} {2b-a} $
So tangents to the curve at points $P$ and $Q$ are parallel.
