# Proof using implicit differentiation

John O' Meara
Prove that if P(a,b) is a point on the rotated ellipse C (whose equation is $$x^2 -xy +y^2=4$$), then so is Q(-a,-b), and that the tangent lines to C at P and Q are parallel.
The equation of the line joining P and Q is
y - b = m(x - a), where $$m= \frac{b-(-b)}{a-(-a)} = \frac{b}{a}$$, then the equation of the line is $$y=\frac{b}{a}x$$. Now, do I substitute for y into the equation for C? To prove the tangent lines are parallel, I have, $$\frac{dy}{dx}=\frac{y-2x}{2y-x}$$.
If I replace x by a/-a and y by b/-b respectively, I get the following for the tangent lines $$\frac{b-2a}{2b-a}, \frac{-b+2a}{-2b+a}$$ which are not the same slope!. I'm studying this on my own so please help. Thanks.

## Answers and Replies

Prove that if P(a,b) is a point on the rotated ellipse C (whose equation is $$x^2 -xy +y^2=4$$), then so is Q(-a,-b), and that the tangent lines to C at P and Q are parallel.
y - b = m(x - a), where $$m= \frac{b-(-b)}{a-(-a)} = \frac{b}{a}$$, then the equation of the line is $$y=\frac{b}{a}x$$. Now, do I substitute for y into the equation for C? To prove the tangent lines are parallel, I have, $$\frac{dy}{dx}=\frac{y-2x}{2y-x}$$.
If I replace x by a/-a and y by b/-b respectively, I get the following for the tangent lines $$\frac{b-2a}{2b-a}, \frac{-b+2a}{-2b+a}$$ which are not the same slope!. I'm studying this on my own so please help. Thanks.