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  • An ellipse inscribed in a semi-circle touches the circular arc at two distant points and also touches the bonding diameter. Its major axis is parallel to the bonding diameter. When the ellipse has

An ellipse inscribed in a semi-circle touches the circular arc at two distant points and also touches the bonding diameter. Its major axis is parallel to the bonding diameter. When the ellipse has the maximum possible area, then its eccentricity is

Option: 1

\frac{1}{\sqrt{2}}


Option: 2

\frac{1}{2}


Option: 3

\frac{1}{\sqrt{3}}


Option: 4

\sqrt{\frac{2}{3}}


Answers (1)

best_answer

Let ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}

and circle \mathrm{x^2+(y+b)^2=r^2 \quad\{ let \, \, radius =r\}}

Put \mathrm{x^2=a^2-\frac{a^2 y^2}{b^2}}

In circle, \mathrm{a^2-\frac{a^2 y^2}{b^2}+(y+b)^2=r^2}

\mathrm{ \begin{aligned} & \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}} \\\\ & \text { Area }=\Delta=\pi a b=\pi a^2 \sqrt{1-\frac{a^2}{r^2}} \\\\ & \frac{d \Delta}{d a}=0 \Rightarrow a^2=\frac{2 r^2}{3} \Rightarrow a=\sqrt{\frac{2}{3}} r \\\\ & \therefore b=a \sqrt{1-\frac{2}{3}}=\frac{a}{\sqrt{3}} \Rightarrow e=\sqrt{\frac{2}{3}} \end{aligned} }

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Ajit Kumar Dubey

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