Two Ellipses, Eccentricity 1/√3: Area of Quadrilateral ABCD

JEE Main 2025 — 29 January, Shift 1. Previous Year Question.

Problem

Let the ellipse $E_1: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ (with $a>b$) and $E_2: \dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1$ (with $A<B$) have the same eccentricity $\dfrac{1}{\sqrt3}$. Let the product of their latus rectum lengths be $\dfrac{32}{\sqrt3}$, and the distance between the foci of $E_1$ be $4$. If $E_1$ and $E_2$ meet at points $A, B, C, D$, then the area of quadrilateral $ABCD$ equals:

Key insight. $E_1$ has its major axis along the $x$-axis and $E_2$ along the $y$-axis, but both share the same eccentricity — which means the same relationship between their semi-axes applies to both, just with the roles of the axes swapped. That symmetry, combined with the two numeric clues (latus rectum product, focal distance), pins down all four unknowns before any intersection points need to be found.

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Approach

Since both ellipses are centred at the origin with perpendicular major axes and the same eccentricity, they intersect in a symmetric pattern — four points forming a rectangle. The strategy is to first pin down $a$, $b$, $A$, $B$ using the two given numeric conditions (same eccentricity gives one relationship per ellipse; the latus rectum product and focal distance give two more equations), then substitute both ellipse equations together to find the intersection points, and finally compute the rectangle’s area directly from those coordinates.

Solution

Step 1 — Express the eccentricity condition for both ellipses

For $E_1$ (major axis along $x$): $e^2 = 1-\dfrac{b^2}{a^2} = \dfrac13 \implies \dfrac{b^2}{a^2}=\dfrac23$.

For $E_2$ (major axis along $y$, so the roles of the semi-axes flip): $e^2 = 1-\dfrac{A^2}{B^2}=\dfrac13 \implies \dfrac{A^2}{B^2}=\dfrac23$.

Step 2 — Use the latus rectum product

The latus rectum length of $E_1$ is $\dfrac{2b^2}{a}$, and of $E_2$ is $\dfrac{2A^2}{B}$. Their product:

$$\frac{2b^2}{a}\cdot\frac{2A^2}{B} = \frac{32}{\sqrt3}$$

Writing $\dfrac{b^2}{a} = \dfrac{b^2}{a^2}\cdot a = \dfrac{2}{3}a$ and $\dfrac{A^2}{B} = \dfrac{A^2}{B^2}\cdot B = \dfrac23 B$:

$$4\cdot\frac{2}{3}a\cdot\frac{2}{3}B = \frac{32}{\sqrt3} \implies \frac{16}{9}aB = \frac{32}{\sqrt3} \implies aB = 6\sqrt3$$

Step 3 — Use the focal distance of $E_1$

The distance between the foci of $E_1$ is $2c$ where $c=ae$. Given $2c=4$, so $c=2$, and since $e=\dfrac{1}{\sqrt3}$:

$$a = \frac{c}{e} = 2\sqrt3$$

Step 4 — Solve for the remaining semi-axes

From $aB=6\sqrt3$ and $a=2\sqrt3$: $B = \dfrac{6\sqrt3}{2\sqrt3}=3$.

From $\dfrac{b^2}{a^2}=\dfrac23$: $b^2 = \dfrac23\times(2\sqrt3)^2 = \dfrac23\times12=8$.

From $\dfrac{A^2}{B^2}=\dfrac23$: $A^2=\dfrac23\times9=6$.

Step 5 — Write out both ellipse equations and find the intersection

$$E_1:\ \frac{x^2}{12}+\frac{y^2}{8}=1, \qquad E_2:\ \frac{x^2}{6}+\frac{y^2}{9}=1$$

Multiplying $E_1$ by $2$ and subtracting $E_2$ to eliminate $x^2$:

$$\frac{2x^2}{12}-\frac{x^2}{6} = 0 \quad\text{(the $x^2$ terms cancel)}, \qquad \frac{2y^2}{8}-\frac{y^2}{9} = 2-1=1$$

$$\frac{y^2}{4}-\frac{y^2}{9}=1 \implies \frac{9y^2-4y^2}{36}=1 \implies y^2 = \frac{36}{5}$$

Substituting back into $E_2$ to find $x^2$:

$$\frac{x^2}{6} = 1 – \frac{y^2}{9} = 1-\frac{36/5}{9} = 1-\frac{4}{5}=\frac15 \implies x^2 = \frac{6}{5}$$

Step 6 — Compute the area of the rectangle

The four intersection points are $\left(\pm\sqrt{\tfrac65},\ \pm\sqrt{\tfrac{36}{5}}\right)$, forming a rectangle with side lengths $2\sqrt{\tfrac65}$ and $2\sqrt{\tfrac{36}{5}}$. The area is:

$$2\sqrt{\frac65}\times 2\sqrt{\frac{36}{5}} = 4\sqrt{\frac{6\times36}{25}} = 4\times\frac{\sqrt{216}}{5} = \frac{4\times 6\sqrt6}{5} = \frac{24\sqrt6}{5}$$

Answer

Common mistakes

• Applying the eccentricity formula the same way to both ellipses without noticing $E_2$’s major axis is along $y$, not $x$ — this flips which semi-axis is “larger” in the formula and gives an inconsistent system if not handled carefully.
• Trying to find the intersection points before pinning down all four constants ($a$, $b$, $A$, $B$) — without first using the eccentricity, latus rectum, and focal distance conditions, the two ellipse equations have too many unknowns to solve directly.

Practise next

• Two ellipses share the eccentricity $\tfrac12$, one with its major axis along the $x$-axis and the other along the $y$-axis. Their latus rectum lengths multiply to $81$, and the distance between the foci of $E_1$ is $6$. Find the area of the quadrilateral formed where they meet, by the same method.