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Eccentricity from Triangle Area Ratio in Ellipse | JEE

JEE Maths question with a full step-by-step solution.

Question

The area of the triangle inscribed in an ellipse bears the ratio

\sqrt5:3

to the area of the triangle formed by joining points on the auxiliary circle corresponding to the vertices of the first triangle. Then the eccentricity of the ellipse is:

\dfrac{2}{3}

correct

\dfrac{3}{4}

\dfrac{\sqrt5}{3}

\dfrac{3}{5}

Solution

Step 1: Relate the two areas. If

P(\theta),Q(\beta),R(\alpha)

are points on the ellipse, the corresponding auxiliary-circle points are obtained by replacing

b\sin

with

a\sin

. Writing both areas as determinants,

\Delta_{\text{ellipse}}=\frac12\,ab\begin{vmatrix}\cos\theta&\sin\theta&1\\ \cos\beta&\sin\beta&1\\ \cos\alpha&\sin\alpha&1\end{vmatrix},\qquad \Delta_{\text{auxiliary}}=\frac12\,a^2\begin{vmatrix}\cos\theta&\sin\theta&1\\ \cos\beta&\sin\beta&1\\ \cos\alpha&\sin\alpha&1\end{vmatrix}.

Step 2: Their ratio is

\frac{\Delta_{\text{ellipse}}}{\Delta_{\text{auxiliary}}}=\frac{b}{a}=\frac{\sqrt5}{3}.

Step 3: Then

e^2=1-\frac{b^2}{a^2}=1-\frac59=\frac49 \;\Rightarrow\; e=\frac23.

Correct answer: (1)

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