Conics (1 Viewer)

[Affinity]

I am quite stuck with this question

P is a variable point on the ellipse x^2/a^2 + y^2/b^2 = 1
S, S' are the Foci of the ellipse. PS, PS' meet the ellipse again at Q, Q' Respectively. The tangents to the ellipse at Q and Q' meet at R. Find the locus of R.

So far I can only get up to

R has paramatric representation:

x=a[cos( (m+n)/2 )/cos( (m-n)/2 )]
y=b[sin( (m+n)/2 )/cos( (m-n)/2 )]

where the following equalities:

e*cos( (m+p)/2 ) = cos( (m-p)/2 )
-e*cos( (n+p)/2 ) = cos( (n-p)/2 )

for some p. ( P = (acos(p), bsin(p) )
(e = eccentricity of ellipse)
holds

and the algebra here looks too daunting

guess = might be an ellipse

any ideas is much appreciated

 

[Affinity]

bashed it, turned out to be an ellipse indeed

x^2/a^2 + y^2/[b(e^2+1)/(e^2-1)]^2 = 1

did anyone find a more *elegant* (easy) way of doing this?

 

[underthesun]

did you go through finding Q and Q', their tangent equations, and the intercept, and put it together?

 

[Affinity]

I did still can't think of a better way to do it.
have you?

 

[Affinity]

What I did was:

Solve simultaneously the equations for tangents at Q and Q' to find R. and found:
R has paramatric representation:

x=a[cos( (m+n)/2 )/cos( (m-n)/2 )]
y=b[sin( (m+n)/2 )/cos( (m-n)/2 )]


Substituted the coordinates of the focus into the equation of the chords PQ, PQ' to obtain

e*cos( (m+p)/2 ) = cos( (m-p)/2 ) (*)
-e*cos( (n+p)/2 ) = cos( (n-p)/2 ) (**)

multiplying * with ** LHS with RHS then solving gives
x= -acos(p)

solving for y was messy
y = -b[(e^2+1)/(e^2-1)]sin(p)

 

[underthesun]

I did that that way too, and it seems that I can't find any easier way to do it.

I gave it up since it seems very messy, and that there could probably be another way to do it.