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Consider an ellipse with major and minor axes of length $10$ and $8$ respectively. The radius of the largest circle that can be inscribed in this ellipse, given that the centre of this circle is one of the focus of the ellipse.

I attempted to solve this by using a very simple concept:

All points of the circle should either lie inside or on the circle. Hence assuming any point on the circle to be $x=ae+r\cos \theta$ and $y=r\sin \theta$ which satisfies:

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}<1$$

I got a quadratic in $\cos \theta$ and I made the equation to be true independent of theta which gave $r\in [a-ae,a+ae]$ what is wrong to solve it by this method.

I would also like to know any other methods to solve this problem.

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Hint:

  1. Parametrize the ellipse as $(x,y) = (5 \cos t, 4 \sin t)$.
  2. Note that a focus is located at $(3,0)$.
  3. Obtain an expression for the distance of a point on the ellipse to this focus.
  4. Find a value of $t$ for which this distance is minimized.
  5. Compute the minimum distance. This is the radius of the largest circle centered at the focus.
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Instead of circle parametrization, write the ellipse in polar coordinates with focus-centred:

$$r=\frac{b^2}{a+c\cos \theta}$$

The maximal interior circle has its radius being the minimal focal distance (i.e. at apogee):

\begin{align} r_{-} &= \frac{b^2}{a+c} \\ &= \frac{b^2}{a+\sqrt{a^2-b^2}} \\ &= \frac{4^2}{5+\sqrt{5^2-4^2}} \\ &= 2 \end{align}

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Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} =1$. Parametric point on the ellipse is $(a\cos t, b\sin t)$ . Distance of centre $(0,0)$ from this point is given by $(a-ae \cos t)$. (Substitute in distance formula and simplify to obtain that expression)

$(a-ae \cos t)$ in this case is $(5-3\cos t)$.

It is minimum for $\cos t=1$ and its value is $2$. Hence , the radius of the largest circle that can be inscribed is $2$ units.

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