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# ellipse problem

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Consider the ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$, where $$a>b>0$$. As a function of $$a$$ and $$b$$, find the radius of the smallest circle that contains the ellipse, is centered on the $$y$$-axis, and which intersects the ellipse only at $$(0,b)$$.

Aug 13, 2022

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Let $$r$$ be the radius of the circle. Then the center of the circle is $$(0,b-r),$$ so the equation of the circle is $$x^2 + (y - (b - r))^2 = r^2.$$
Solving for $$x^2$$ in $$\frac{x^2}{a^2}+\frac{y^2}{b^2},$$ we get $$x^2 = \frac{a^2 (b^2 - y^2)}{b^2}.$$
Substituting this into $$x^2 + (y - (b - r))^2 = r^2,$$ we get $$\frac{a^2 (b^2 - y^2)}{b^2} + (y - (b - r))^2 = r^2.$$
This simplifies to $$(a^2 - b^2) y^2 + (2b^3 - 2b^2 r) y + (2b^3 r - a^2 b^2 - b^4) = 0.$$
Since the ellipse and circle intersect at $$(0,b),$$ $$y = b$$ will be a root of the quadratic above. Thus, we can take out a factor of $$y - b,$$ which gives us $$(y - b)((a^2 - b^2) y + a^2 b + b^3 - 2b^2 r) = 0.$$
The solution to $$(y - b)((a^2 - b^2) y + a^2 b + b^3 - 2b^2 r) = 0$$ is $$y_1 = \frac{2b^2 r - a^2 b - b^3}{a^2 - b^2}.$$
If $$y_1 < b,$$ then the ellipse and circle also intersect at a point where the $$y$$-coordinate is $$y_1$$ (or the circle does not contain the ellipse). So, we need $$y_1 = \frac{2b^2 r - a^2 b - b^3}{a^2 - b^2} \ge b.$$
Isolating $$r,$$ we find $$r \ge \frac{a^2}{b}.$$
Therefore, the smallest possible radius is $$\boxed{\frac{a^2}{b}}.$$

Aug 18, 2022
edited by WorldEndSymphony  Aug 18, 2022