Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 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} }

Posted by

Ajit Kumar Dubey

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: NIT Tiruchirappalli cut-offs rose for major BTech programmes this year; closing ranks
anam-khan

JoSAA 2025 reconstituted; NIT Rourkela to coordinate BTech counselling for IITs, NITs, IIITs
anam-khan

JEE Main 2025: BTech Information Technology cut-offs at NITs; know opening, closing ranks

Latest Question