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From a point perpendicular tangents are drawn on the ellipse $x^2+2y^2=2$. The chord of contact touches a circle concentric with ellipse. Find ratio of min and max area of circle

Let the point from which tangents are drawn be $(h,k)$

Then the locus of that point will be $$h^2+k^2=3$$

Also the chord of contact is $$\frac{hx}{2}+ky-(\frac {h^2}{2}+k^2)=0$$

Let the circle be $$x^2+y^2=a^2$$

Then the tangent to this circle is $$y=\frac{-h}{2k}x\pm a\sqrt{1+\frac{h^2}{4k^2}}$$

$$hx+2ky \mp a\sqrt{4k^2+h^2}=0$$ Now I could equate the $c$ term of the linear equations, but that’s a very lengthy process, so I am convinced I am approaching the question wrong. How should I do it right?

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    $\begingroup$ It seems likely that the extrema are attained when $h=0$ or $k=0$. I suspect that the envelope of these polar lines is an ellipse, so I’d see if I can form the dual to it from the equations of those polars. $\endgroup$
    – amd
    May 29 '20 at 19:16
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The correct equation of the chord of contact is: $$ \frac{hx}{2}+ky-\frac {h^2+k^2}{3}=0 $$ and its distance from the origin (i.e. the radius of the tangent circle) is thus $$ r={2\over3}{h^2+k^2\over \sqrt{h^2+4k^2}}={2\over\sqrt3}{1\over \sqrt{1+k^2}}. $$ From that, it's easy to find minimum and maximum value of $r$.

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  • $\begingroup$ How is that the right equation for the chord of contact? Isn’t it $T=S_1$ $\endgroup$
    – Aditya
    May 30 '20 at 6:31
  • $\begingroup$ The chord of contact is the line through the tangency points. Your line passes through $(h,k)$. $\endgroup$ May 30 '20 at 8:37
  • $\begingroup$ $(h,k)$ aren’t the tangency points though $\endgroup$
    – Aditya
    May 30 '20 at 14:06
  • $\begingroup$ Probably I wasn't clear enough: YOUR line passes through point $(h,k)$ and not through tangency points (please check it!), MY line passes through tangency points and not through point $(h,k)$. $\endgroup$ May 30 '20 at 14:34
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    $\begingroup$ The minimum is for $k^2=3$, obviously. $\endgroup$ May 31 '20 at 6:24
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If we parameterize the circle as $(h,k)=(\sqrt3\cos t,\sqrt3\sin t)$, we then have the one-parameter family of polar lines $$x\sqrt3\cos t+2y\sqrt3\sin t-2=0.$$ The square of the distance of this line to the origin is $${4\over3(\cos^2t+4\sin^2t)} = {4\over3(1+3\sin^2 t)},$$ which has extrema at $t=0$ and $t=\pi/2$, yielding max/min distances of $2/\sqrt3$ and $1/\sqrt3$.

We can go a bit further and compute the envelope of those polar lines. Taking the generic equation of a line $\lambda x+\mu y+\tau = 0$, equating coefficients and eliminating $t$, we get the conic equation $$4\lambda^2+\mu^2-3\tau^2=0.$$ This is dual to the ellipse $$\frac{x^2}4+y^2=\frac13,$$ from which we read the semiaxis lengths $\frac2{\sqrt3}$ and $\frac1{\sqrt3}$.

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  • $\begingroup$ So if we figure out at what points extrema and minima land at, then there is no need for further calculations $\endgroup$
    – Aditya
    May 30 '20 at 6:44
  • $\begingroup$ @Aditya Well, since the problem asks for the ratio of the largest and smallest circles, there’s still the ratios of the squares of these two distances to compute, but otherwise you’re done. That ratio happens to equal $1-e^2$ for the above elliptical envelope, but I can’t think of a good way to compute that directly that doesn’t involve first computing its semiaxis lengths. $\endgroup$
    – amd
    May 30 '20 at 18:06

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