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An ellipse inscribed in a semi-circle touches the circular arc at two distinct points and also touches the bounding diameter. Its major axis is parallel to the bounding diameter. When the ellipse has the maximum possible area, its eccentricity is
$\cdot \frac{1}{\sqrt{2}}$
$\frac{1}{2}$
$.. \frac{1}{\sqrt{3}}$
$\sqrt{\frac{2}{3}}$
Solution
(d)
Let equation of ellipse is
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\therefore$ Equation of circle is
$x^2+(y+b)^2=r^2$
Put $x^2=a^2-\frac{a^2 y^2}{b^2}$ in circle
$a^2-\frac{a^2 y^2}{b^2}+(y+b)^2=r^2$
$\Rightarrow\left(1-\frac{a^2}{b^2}\right) y^2+2 b y+\left(a^2+b^2-r^2\right)=0$
$D=0 \Rightarrow r^2=\frac{a^4}{a^2-b^2} \Rightarrow b=a \sqrt{1-\frac{a^2}{r^2}}$
Area of ellipse $=\pi a b$
$A=\pi a^2 \sqrt{1-\frac{a^2}{r^2}}$
$\frac{d A}{d a}=0 \Rightarrow a^2=\frac{2 r^2}{3} \Rightarrow a=\sqrt{\frac{2}{3} r}$
$\therefore \quad b=a \sqrt{1-\frac{2}{3}}=\frac{a}{\sqrt{3}}$
$\Rightarrow \quad e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{1}{3}}=\sqrt{\frac{2}{3}}$
