JEE Main 2025 — 22 January, Evening Shift. Previous Year Question.
Problem
Let $E: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ ($a>b$) and $H: \dfrac{x^2}{A^2}-\dfrac{y^2}{B^2}=1$. Let the distance between the foci of $E$ and the distance between the foci of $H$ both equal $2\sqrt3$. If $a-A=2$, and the ratio of the eccentricities of $E$ and $H$ is $\dfrac{1}{3}$, then the sum of the lengths of their latus rectums is equal to:
Key insight. “Distance between foci” for both curves equals $2\sqrt3$ — meaning $ae_1 = Ae_2 = \sqrt3$ for the ellipse’s eccentricity $e_1$ and the hyperbola’s eccentricity $e_2$. Combined with the given ratio $e_1:e_2=1:3$, this turns into a system that pins down $a$ and $A$ individually — after which $b^2$ and $B^2$ (needed for the latus rectums) follow from the standard eccentricity formulas for each conic.
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Approach
Write the focal-distance condition as $ae_1=\sqrt3$ (for the ellipse) and $Ae_2=\sqrt3$ (for the hyperbola). Use $e_2=3e_1$ (from the given ratio) to relate $a$ and $A$, then combine with $a-A=2$ to solve for both. Once $a$, $A$, $e_1$, $e_2$ are known, find $b^2=a^2(1-e_1^2)$ and $B^2=A^2(e_2^2-1)$, and compute each latus rectum as $\frac{2b^2}{a}$ and $\frac{2B^2}{A}$.
Solution
Step 1 — Write the focal-distance conditions
For the ellipse, distance between foci $=2ae_1=2\sqrt3 \implies ae_1=\sqrt3$.
For the hyperbola, distance between foci $=2Ae_2=2\sqrt3 \implies Ae_2=\sqrt3$.
Step 2 — Use the eccentricity ratio
Given $\dfrac{e_1}{e_2}=\dfrac{1}{3}$, so $e_2=3e_1$. Substituting into $Ae_2=\sqrt3$:
$$A(3e_1)=\sqrt3 \implies Ae_1 = \frac{\sqrt3}{3} = \frac{1}{\sqrt3}$$
Step 3 — Find the ratio a/A
Dividing $ae_1=\sqrt3$ by $Ae_1=\frac{1}{\sqrt3}$:
$$\frac{a}{A} = \frac{\sqrt3}{\frac{1}{\sqrt3}} = 3 \implies a=3A$$
Step 4 — Solve for a and A using a − A = 2
$$3A-A=2 \implies A=1, \qquad a=3$$
Step 5 — Find the eccentricities
From $ae_1=\sqrt3$: $e_1 = \dfrac{\sqrt3}{3} = \dfrac{1}{\sqrt3} \implies e_1^2=\dfrac{1}{3}$
$$e_2 = 3e_1 = \sqrt3 \implies e_2^2=3$$
Step 6 — Find b² and B²
For the ellipse: $b^2=a^2(1-e_1^2) = 9\left(1-\frac{1}{3}\right) = 9\times\frac{2}{3}=6$
For the hyperbola: $B^2=A^2(e_2^2-1) = 1\times(3-1)=2$
Step 7 — Compute each latus rectum
$$l_E = \frac{2b^2}{a} = \frac{2(6)}{3}=4, \qquad l_H = \frac{2B^2}{A} = \frac{2(2)}{1}=4$$
Step 8 — Sum them
$$l_E+l_H = 4+4=8$$
Answer
$$8$$
Common mistakes
- Confusing “distance between the two foci” (which is $2ae$) with “distance from centre to a focus” ($ae$). Using $ae=2\sqrt3$ instead of $ae=\sqrt3$ doubles every subsequent value incorrectly.
- Using $b^2=a^2+e^2$ or a similarly malformed identity instead of the correct $b^2=a^2(1-e^2)$ for an ellipse and $B^2=A^2(e^2-1)$ for a hyperbola — these two formulas have a crucial sign difference that’s easy to mix up.
Practise next
- An ellipse and a hyperbola share the same foci, $2\sqrt5$ apart. If $a-A=2$ and their eccentricities are in the ratio $1:2$, find the sum of their latus rectum lengths, using the same approach.
Show answer
$\dfrac{13}{2}$. Both conics have $c=\sqrt5$, so $a=\dfrac{\sqrt5}{e_1}$ and $A=\dfrac{\sqrt5}{e_2}=\dfrac{\sqrt5}{2e_1}$.
Then $a-A=\dfrac{\sqrt5}{2e_1}=2$ gives $e_1=\dfrac{\sqrt5}{4}$, so $a=4$ and $A=2$ — and $e_1<1<e_2$, as an ellipse and a hyperbola require.
Hence $b^2=16-5=11$ and $B^2=5-4=1$, so the latus recta are $\dfrac{2b^2}{a}=\dfrac{11}{2}$ and $\dfrac{2B^2}{A}=1$, summing to $\dfrac{13}{2}$.
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.