Parametrics HELP (1 Viewer)

[bored.of.u]

Can u guys help me with this parametric question. Tried a few ways of doing it but cannot get the answer:

Q: P is a variable point (x1, y1) on the parabola x^2 = 4ay with focus S. The tangent at P meets the tangent at the vertex of the parabola at Q and it meets the axis of the parabola at R. Show SQ is perpendicular to PQ.

thnks guys

 

[addikaye03]

I haven't tried it yet, i will tomorrow morning for you but two approaches i've thought of:

- m1m2=-1

-Find an expression for the distance of each and use phythagoras to show that its 90 degree.

-Perpendicular distance (possibly)

You probably have already thought of these but just thought i would put them out there... I'll do it tomoz morning for you

 

[scardizzle]

okay here's my attempt:

differentiating 4ay = x^2

gives us y'= x/2a

therefore gradient at x = x1 is x1/2a

now to work out the gradient of SQ

since we know S is (0,a)

we must find out Q

Eq'n of the tangent at P is:

2a(y-y1)=x1(x-x1)

for the point Q y must = 0 since it intersect the x-axis (tangent of the vertex)

therefore -2ay1 = x1(x-x1) ---1

we also know the x1^2 = 4ay1

therefore y1 = x1^2/4a

subbing into into 1:

-x1^2/2 = x1(x-x1)

which gives us x = x1/2

using the gradient formula for the gradient of SQ

a/(-x1/2) = -2a/x1

now m of P x m of SQ = -1

therefore lines are perpentdicular

please correct me if ive made any mistakes