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Orthogonal Tangents (Posted on 2011-11-30) Difficulty: 3 of 5
Let Γ be a parabola with focus F and directrix d.
A line through F intersects Γ in points P1 and P2.
The feet of the perpendiculars from P1 and P2 on
d are Q1 and Q2 respectively.
The midpoint of line segment Q1Q2 is Q0.

Prove that the rays Q0P1 and Q0P2 are orthogonal
and that they are tangent to Γ.

See The Solution Submitted by Bractals    
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solution | Comment 1 of 2

without loss of generality:
let the vertex be at the origin,
let the directrix be y=-p
let the focus be (0,p)
then the parabola is given by
x^2=4py

now let P1 be (2pk,pk^2) for some real number k
then the line P1-F is given by
y-pk^2=(pk^2-p)(x-2pk)/(2pk)
solving for the other intersection gives
P2 at (-2p/k,p/k^2)

thus the feet Q1,Q2 are
Q1: (2pk,-p)
Q2: (-2p/k,-p)
thus
Q0: (p(k^2-1)/k,-p)
thus the rays Q0P1 and Q0P2 have slopes
Q0P1: m1=k
Q0P2: m2=-1/k
m1*m2=k*-1/k=-1
thus Q0P1 and Q0P2 are orthogonal


  Posted by Daniel on 2011-12-01 10:35:19
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