How Do I Solve a Complex Coordinate Geometry Problem with Parabolas?

[chickens]

Hi there, I've been struggling with this question for days , the first part where they call me to prove that equation, i could do it...the second one i don't know how to do...could anyone help me? how do i find the locus equation? its so confusing... ty in advance.


If the normal at P(ap^2, 2ap) to the parabola y^2=4ax meets the curve again at Q(aq^2,2aq), prove that p^2 + pq + 2 = 0. Prove that the equation of the locus of the point of intersection of the tangents at P and Q to the parabola is y^2(x + 2a) + 4a^3 = 0.

^ means to the power of...

 

[devious_]

I know I'm supposed to give hints not complete solutions, but I really didn't know what kind of hint to give...

The equation of the tangent to the parabola at the point (at^2, 2at) is:
y = (1/t)x + at (I'll leave the proof for you.)

Using this equation for points P and Q, we get:
y = (1/p)x + ap
y = (1/q)x + aq

Solving these simultaneously twice (once for x and once for y), we get:
(1) pq = x/a
(2) p+q = y/a

Re-writing (1) as q=x/(ap), then using it in (2) gives us:
p^2 = (py-x)/a

Now using p^2+pq+2=0 with the values for p^2 and q we just obtained gives us:
(py-x)/a + x/a + 2 = 0
p = (-2a)/y

Now re-writing (2) as q=(y^2+2a)/ay and using the value of p we just obtained in (1) yields:
pq = (-2y^2 - 4a^2)/y^2 = x/a

Upon multiplying it out we get:
-2ay^2 - 4a^3 = xy^2

Rearranging:
xy^2 + 2ay^2 + 4a^3 = 0
y^2(x+2a) + 4a^3 = 0, as required.