TITLE: Find the latus rectum of the Parabola
QUESTION [9 upvotes]: Let $y=3x-8$ be the equation of tangent at the point $(7,13)$ lying on a parabola, whose focus is at $(-1,-1)$. Evaluate the length of the latus rectum of the parabola.  

I got this question in my weekly test. I tried to assume the general equation of the parabola and solve the system of equations to calculate the coefficients with the help of these given conditions. But this way it becomes very lengthy and tedious. Can anyone provide an elegant solution? Thanks.

REPLY [5 votes]: Denote:
F: Focus (-1,-1)
P: The point on the parabola (7,13)
M: the foot of the perpendicular from the Focus F on the tangent at P
V: Vertex of Parabola
Then a standard property of the parabola (Prop 125 in Askwith - A Course of Pure Geometry) states that $FM^2 = FP.FV$.
We easily obtain $FM = \sqrt {10}$ and $FP = 2 \sqrt{65}$
Then $FV = \sqrt{\frac{5}{13}}$. 
The length of latus rectum $=4 FV = 4 \sqrt{\frac{5}{13}}$