If tangents to thew parabola y^2=4ax at the point P &Q intersect at T.?

Question

Prove that Tp and TQ subtend equal angles at the focus

Answer 1

This one is tough. I've looked at it three different ways and still don't get it.

First try. If P(x1, y1) and Q(x2, y2) are given, and y^2 = 4ax, then the derivative is 2y dy = 4a dx ==> m = dy/dx = 2a/y.

So (y - y1) / (x - x1) = m1 = 2a/y1. Solve for y to get

y = (2a/y1) (x - x1) + y1 (equation of a tangent line)

Repeat for (x2, y2), then set the two y-values equal to each other, and solve the resulting equation for x at the point where the two lines intersect. That result is

x = [y2 (y1y2 + 2ax1) - y1 (y1y2 + 2ax2)] / [2a(y2 - y1)]

so it's possible to find the point of intersection T, but the algebra is very messy. That's where I stopped.

Second try. Parallel rays of sunlight striking a parabolic mirror will get reflected off the tangent and redirected to the focus. Snell's law says the angle of incidence equals the angle of reflection. That gave me a lot of angle relationships, but I didn't make much progress.

Third try. I got it to work for one special case. Let P(a,2a) (the top end of the latus rectum) and Q(4a, -4a) be given. Then the slope of the tangent is given by m = 2a/y, and

m1 = 2a/y1 = 1 and m2 = 2a/y2 = -1/2

From this we find that the two tangent lines intersect at T(-2a,-a).

To get the angles POT and QOT (where O is the focus at (a,0)), I got three vectors OP = <2aj>; OQ = <3ai - 4aj>; and OT = <-3ai - aj>. The magnitudes are OP = 2a, OQ = 5a, OT = a sqrt(10). Then using the inner (dot) product:

OP dot OT = <2aj> dot <-3ai - aj> = -2a^2 = 2a^2 sqrt(10) cos POT, so cos POT = -1/sqrt(10) and angle POT = -108.435 degrees.

OQ dot OT = <3ai - 4aj> dot <-3ai - aj> = -5a^2 = 5a^2 sqrt(10) cos (QOT), so cos QOT = -1/sqrt(10) and angle QOT = -108.435 degrees.

In this special case, the angles subtended by PT and QT are equal.

And that's better than nothing.

Answer 2

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