JEE Main 2024. Previous Year Question.
Problem
If the length of the minor axis of an ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is:
Key insight. The given condition directly relates $b$ (half the minor axis) to $c$ (half the distance between foci) — translating it into $c=2b$ pins down the ratio between them immediately, and the standard relation $a^2=b^2+c^2$ does the rest.
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Approach
Write the length of the minor axis as $2b$ and the distance between the foci as $2c$. Translate the given condition into an equation relating $b$ and $c$. Use the standard relation $a^2=b^2+c^2$ (with $b$ and $c$ now both expressed in terms of one variable) to find $c^2/a^2$, and take the square root to get the eccentricity $e=c/a$.
Solution
Step 1 — Translate the given condition
Length of minor axis $=2b$; distance between foci $=2c$. The condition states:
$$2b = \frac{1}{2}(2c) = c$$
So $c=2b$.
Step 2 — Use the standard ellipse relation
$$a^2 = b^2+c^2$$
Step 3 — Substitute c = 2b
$$a^2 = b^2+(2b)^2 = b^2+4b^2 = 5b^2$$
Step 4 — Find c²/a²
$$\frac{c^2}{a^2} = \frac{(2b)^2}{5b^2} = \frac{4b^2}{5b^2} = \frac{4}{5}$$
Step 5 — Take the square root to find the eccentricity
$$e = \frac{c}{a} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt5}$$
Answer
$$e = \frac{2}{\sqrt5}$$
Common mistakes
- Confusing the minor axis length ($2b$) with the semi-minor axis ($b$), or the distance between foci ($2c$) with $c$ itself. The problem’s condition is stated in terms of the full lengths (axis length, full focal distance) — carefully converting to the standard $b$ and $c$ notation before substituting into $a^2=b^2+c^2$ avoids a factor-of-2 error.
- Forgetting to take the square root at the end. Computing $c^2/a^2=4/5$ gives the square of the eccentricity — the eccentricity itself is $\sqrt{4/5}=2/\sqrt5$, not $4/5$.
Practise next
- If the length of the major axis of an ellipse is equal to three times the distance between the foci, find the eccentricity, using the same substitution approach with $a^2=b^2+c^2$.
Show answer
$e=\dfrac13$. The major axis is $2a$ and the distance between the foci is $2ae$.
So $2a=3\cdot2ae$, giving $e=\tfrac13$ directly — $a$ cancels, which it must, since eccentricity describes shape, not size.
For completeness, $b^2=a^2(1-e^2)=\tfrac89a^2$, so the minor axis is $\tfrac{2\sqrt2}{3}$ of the major.
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