Prove that if P(a,b) is a point on the rotated ellipse C (whose equation is [tex] x^2 -xy +y^2=4 [/tex]), then so is Q(-a,-b), and that the tangent lines to C at P and Q are parallel.(adsbygoogle = window.adsbygoogle || []).push({});

The equation of the line joining P and Q is

y - b = m(x - a), where [tex] m= \frac{b-(-b)}{a-(-a)} = \frac{b}{a}[/tex], then the equation of the line is [tex] y=\frac{b}{a}x [/tex]. Now, do I substitute for y into the equation for C? To prove the tangent lines are parallel, I have, [tex] \frac{dy}{dx}=\frac{y-2x}{2y-x}[/tex].

If I replace x by a/-a and y by b/-b respectively, I get the following for the tangent lines [tex] \frac{b-2a}{2b-a}, \frac{-b+2a}{-2b+a}[/tex] which are not the same slope!. I'm studying this on my own so please help. Thanks.

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# Proof using implicit differentiation

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