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...

Question

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?

Answer 1

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 $r$.

Answer 2

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 $\frac1{\sqrt3}$.