Hi guys!

I am baffled by my own solution to the following parabola question. Having got the right answer, I find that I don't like it! Can anyone help?

The parabola is $\displaystyle y^2 = 4ax$ and at point $\displaystyle P(ap^2, 2ap)$ the tangent is found to be $\displaystyle yp = x + ap^2$. We have to find the coordinates of the point of intersection T of the tangent at P and the tangent at $\displaystyle Q(ap^2, 2aq)$.

In the derivation of the tangent at P the gradient is found to be 1/p.

To get the tangent at Q, given that P and Q have the same coordinate $\displaystyle ap^2$, I would expect the intersection to be on the x-axis (since $\displaystyle y^2 = 4ax$ is symmetrical). The answer does not confirm this, so that's my first question - why not?

If I put the coords of Q into the parabola I get:

$\displaystyle (2aq)^2 = 4a(ap^2)$

$\displaystyle q^2 = p^2$

$\displaystyle q = \pm p$

I therefore take the gradient at Q to be -1/p, then the equation of the tangent follows:

$\displaystyle \frac{y - 2aq)}{x - ap^2} = \frac{-1}{p}$

call this equation 1

$\displaystyle yp - 2apq = -x + ap^2$

The tangent at P is

$\displaystyle yp = x + ap^2$

subtracting gives

$\displaystyle x = apq$

$\displaystyle y = a(p + q)$

which is the correct answer.

The thing that puzzles me is that to get the answer I am assuming the gradient at Q is symmetrical i.e. -1/p (since the gradient at P is 1/p). Also if I express this gradient as 1/q which is the same thing and substitute into equation 1, I get:

$\displaystyle \frac{y - 2aq)}{x - ap^2} = \frac{1}{q}$

$\displaystyle yq - 2aq^2 = x - ap^2$

and equation 1 is:

$\displaystyle yp = x + ap^2$

subtracting:

$\displaystyle y(p - q) - 2aq^2 = -2ap^2$

$\displaystyle y(p - q) = 2a(q + p)(q - p) $

$\displaystyle y = 2a(q + p)$

A different result! All these ps and qs are swimming before my eyes. Have I made a slip somewhere?