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.

 

[HallsofIvy]

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.

I take you have proven that "if P(a,b) is on C then so is Q(-a,-b)". That's simple substitution.
But I have no idea why you are looking at "the line joining P and Q". That has nothing to do with the tangent lines.
Yes, 2x- y- xy'+ 2yy'= 0 so (2y-x)y'= y- 2x and y'= (y- 2x)/(2y-x). If you replace x by a and y by b, you get (b-2a)/(2b-a) as slope of the tangent line. If you replace x by -a and y by -b, you get (-b+2a)/(-2b+ a) as slope- and those are exactly the same: factor a (-1) out of both numerator and denominator in the second fraction.