Question

For the parabola y^2=4px, where p, the focal length, is a constant, the focal point, f(p,0), is on the x-axis. Point S is 10 units from the point F and the segment SF makes a 60degree angle with the x-axis. The slope of segment TF is twice the negative of the slope of the tangent to the curve at point T. The length of segment TF is ____ units.

There's a picture, and I don't know if this information is needed but the parabola opens to the right with the vertex at the origin.

Answer 1

Ok. The information for S allows us to determine p, since the distance SF equals the distance from S to the line x = -p. So,

2p+5 = 10
p = 5/2

The equation of the parabola is

y^2 = 4(5/2)x = 10x
y = √(10x)

Now we have to look at T. The slope of the tangent to the curve at (x,y) is

dy/dx = 5/√(10x)
So, we want the point (x,y) where

√(10x)/(x-5/2) = (-2)*5/√(10x)
x = 5/4

So, F is the point (5/4,5/√2)

Tangent there has slope √2
TF has slope -2√2