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Suppose that C is any circle concentric with the ellipse E: x^2/16+ y^2/4 =1 . Let A is a point on E and B is a point on C such that AB is tangent to both E and C. The maximum Length of AB is?

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  • $\begingroup$ Also mention what have you done so far in this problem? $\endgroup$
    – Anurag A
    Commented Nov 14, 2017 at 18:59
  • $\begingroup$ I do not know how to proceed. Please send me whole solution if you know. $\endgroup$ Commented Nov 14, 2017 at 19:01

3 Answers 3

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Segments

Let C be $\frac{x^2}{r^2}+\frac{y^2}{r^2}=1, 2\leq r\leq 4$. The dual conics to $E$ and $C$ are $16X^2+4Y^2=1$ and $r^2X^2+r^2Y^2=1$, respectively. They intersect in $(X,Y)=(\pm\frac{\sqrt{r^2-4}}{2\sqrt{3}r},\pm\frac{\sqrt{16-r^2}}{2\sqrt{3}r})$, that is the lines $\pm\frac{\sqrt{r^2-4}}{2\sqrt{3}r} x\pm\frac{\sqrt{16-r^2}}{2\sqrt{3}r} y+1=0$ are the common tangents, where the $\pm$s are taken independently. For more on the intersect-the-dual-conics approach to common tangents see this answer.

The length of the four segments are the same by symmetry so let's focus on the one in the first quadrant. The tangent points $B$ and $A$ are $(\frac{ r\sqrt{r^2-4}}{2\sqrt{3}}, \frac{r\sqrt{16-r^2}}{2\sqrt{3}})$ and $(\frac{8\sqrt{r^2-4}}{\sqrt{3}r},\frac{2\sqrt{16-r^2}}{\sqrt{3}r})$. So the length of $AB$ reduces to $$\sqrt{\frac{-r^4+20r^2-64}{r^2}}.$$ Now it's a calculus exercise to show that the maximum is $2$ for $r=\sqrt{8}$.

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Another way of solving the problem is to take parametric coordinates as trigonometric ratios. For the ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$, $A$ can be written as $(4cos (\theta), 2sin (\theta))$ and for the circle $x^2+y^2=r^2$ with unknown radius $r$, we have $B\equiv(r \cos(\omega), r \sin(\omega))$. Do note that the angles are different.

Now we write the equation of tangent $AB$ in forms for both the ellipse and circle: $$\frac{x \cos\theta}{4}+\frac{y \sin\theta}{2}=1$$ $$x \cos\omega+y \sin\omega=r$$ Since these represent the same line, $\cos\omega=\frac{r}{4}\cos\theta$ and $\sin\omega = \frac{r}{2}\sin\theta$.

Writing the expression for distance $AB$ ($d$): $$d^2 = (r \cos\omega - 4 \cos\theta)^2+(r \sin\omega - 2 \sin\theta)^2$$ which reduces to (by substituting $\cos\omega$ and $\sin\omega$ with $\cos\theta$ and $\sin\theta$): $$d^2 = 12\cos^2\theta + 4 - r^2$$

Now because the point $(r \cos\omega, r \sin\omega)\equiv(\frac{r^2}{4}\cos\theta, \frac{r^2}{2}\sin\theta)$ lies on the circle, we can say that: $$\frac{r^4}{16}\cos^2\theta + \frac{r^4}{4}\sin^2\theta = r^2$$ which reduces to $$r^2 = \frac{16}{3 \sin^2\theta + 1}$$

Now we can either replace $\cos\theta$ (which we can get from simple trigonometric manipulation of the fraction above) in the expression for $d^2$ and maximise it as a function of $r$. Or we could do it in terms of $\cos\theta$ and $\sin\theta$.

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Let's solve this using the slope form of tangents. The equation of tangent to ellipse: $$y=mx\pm {\sqrt{ 16m^2+4}}$$ This is also tangent to the circle of radius $r$ and then the perpendicular distance from the origin to tangent is $r$ $$r=\frac{\vert\sqrt {(16m^2+4)}\vert}{\sqrt{1+m^2}}$$ $$r^2=\frac{(16m^2+4)}{1+m^2}$$ The point of intersection of the tangent to the ellipse is given by: $$(x', y')=(\frac{\pm16m}{\sqrt{16m^2+4}},\frac{\pm4}{\sqrt{16m^2+4}})$$ Now the length of the tangent between the circle and the ellipse is given by $$L=\sqrt{x'^2+y'^2-r^2}$$ Now when we put all the values of x', y' and r we will get an equation in $m$ (slope) only. For maximum length, $\frac{dL}{dm}= 0$. $$\frac{d}{dm}\sqrt{\frac {256m^2}{16m^2+4}+\frac{16}{16m^2+4}-\frac{16m^2+4}{1+m^2}}=0$$ Putting this condition and solving, we get $$m=\pm \frac { 1}{\sqrt2}$$ And then put it back in the length equation to get AB as 2.

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