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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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The answer, for anyone interested,

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

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