The ellipse with equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ has one of its foci at the point $F$. The perpendicular from the origin to the tangent at a point $P(a\cos\theta, b\sin\theta)$ on the ellipse intersects the line $FP$ at a point $G$. Find the locus of $G$ as $\theta$ varies.


HINT - modified:

I would say, the locus will circle:

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$FP:\,y=\frac{b\sin \theta}{a\cos \theta - e}\,(x+e)$

$SR:\,y=\frac{a}{b}\,\tan \theta\, x=\frac{a}{b}\,\tan \theta\,(x+e)-\frac{a\,e}{b}\,\tan \theta$

$\Rightarrow x+e = a\,\frac{a \cos \theta-e}{e\cos \theta -a}, \quad y = a\,\frac{b \sin \theta}{e\cos \theta -a}$


$(x+e^2)+y^2=a^2\left(\frac{(a\cos \theta -e)^2+b^2\sin^2 \theta}{(e\cos \theta -a)^2}\right)=a^2\left(\frac{a^2\cos^2 \theta -2ae\cos \theta +e^2+(a^2 -e^2)\sin^2 \theta}{(e\cos \theta -a)^2}\right)=$

$=a^2\left(\frac{a^2(\cos^2 \theta +\sin^2 \theta)-2ae\cos \theta +e^2\cos^2 \theta}{(e\cos \theta -a)^2}\right)=a^2\left(\frac{(a-e\cos \theta)^2}{(e\cos \theta -a)^2}\right)=a^2$

$\Rightarrow\,$the equation of the locus: $\,(x+e^2)+y^2=a^2$

  • $\begingroup$ I guessed that, but the algebraic working becomes very messy. I got the lines to have equations $y = \frac{a}{b}\tan\theta x$ and $y = \frac{b\sin\theta}{a\cos\theta-f}(x-f)$ where f is the x-coordinate of the focus. I then combined the two to get everything in terms of x and y, but can't seem to get the answer. There must be an easier way than this! Can anyone post some working? $\endgroup$ – wrb98 Nov 11 '15 at 18:55
  • $\begingroup$ @wrb98 The answer is edited. $\endgroup$ – georg Nov 13 '15 at 10:57

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