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Proof using implicit differentiation

  1. Dec 5, 2009 #1
    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.
    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.
  2. jcsd
  3. Dec 6, 2009 #2


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    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.
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