JEE Main 2024. Previous Year Question.
Problem
Let an ellipse with centre $(1,0)$ and latus rectum of length $\dfrac{1}{2}$ have its major axis along the x-axis. If its minor axis subtends an angle of $60°$ at the foci, then the square of the sum of the lengths of its minor and major axes is equal to ____.
Key insight. For any ellipse, the distance from a focus to an endpoint of the minor axis is always exactly $a$ (the semi-major axis) — a direct consequence of the definition of an ellipse. This turns the “angle subtended at the foci” condition into a simple right-triangle relationship between $a$, $b$, and $c$, without needing any trigonometric identities beyond a basic tangent ratio.
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Approach
Since the ellipse is shifted to centre $(1,0)$, treat it exactly like a standard ellipse but with all coordinates shifted by $1$ in the $x$-direction. Use the right triangle formed by a focus, the centre, and an endpoint of the minor axis: the angle at the focus (half of $60°$) gives $\tan30°=b/c$. Combine this with the latus rectum condition and the standard relation $b^2=a^2-c^2$ to solve for $a$ and $b$.
Solution
Step 1 — Set up the right triangle at a focus
At focus $F$ (a distance $c$ from the centre along the major axis), the two endpoints of the minor axis are each at distance $a$ from $F$ (a standard ellipse property). By symmetry, the angle between $F$-to-centre and $F$-to-each endpoint is half of $60°$, i.e. $30°$.
In the right triangle with legs $c$ (horizontal, centre to focus) and $b$ (vertical, centre to minor axis endpoint):
$$\tan30° = \frac{b}{c} \implies \frac{1}{\sqrt3} = \frac{b}{c} \implies c = b\sqrt3$$
Step 2 — Apply the latus rectum condition
$$\frac{2b^2}{a} = \frac{1}{2} \implies b^2 = \frac{a}{4}$$
Step 3 — Use the standard ellipse relation
$$b^2 = a^2-c^2$$
Substituting $c=b\sqrt3$ (so $c^2=3b^2$):
$$b^2 = a^2-3b^2 \implies 4b^2=a^2$$
Step 4 — Substitute b² = a/4 from Step 2
$$4\left(\frac{a}{4}\right) = a^2 \implies a = a^2 \implies a^2-a=0 \implies a(a-1)=0$$
Since $a=0$ is not valid for an ellipse, $a=1$.
Step 5 — Find b
$$b^2 = \frac{a}{4} = \frac{1}{4} \implies b=\frac{1}{2}$$
Step 6 — Compute (2a+2b)²
$$2a+2b = 2(1)+2\left(\frac{1}{2}\right) = 2+1 = 3$$
$$(2a+2b)^2 = 3^2 = 9$$
Answer
$$9$$
Common mistakes
- Forgetting that the distance from a focus to a minor-axis endpoint is $a$, not $b$ or $c$. This is the single fact that makes the whole right-triangle setup work — without it, there’s no clean way to relate the subtended angle to $b$ and $c$.
- Getting confused by the shifted centre $(1,0)$. Since every relevant length (the latus rectum, the axis lengths, the focus-to-vertex distances) is a difference between coordinates, the shift by $1$ unit cancels out entirely and never needs to be tracked through the calculation.
Practise next
- An ellipse centred at $(2,0)$ has latus rectum $\dfrac{1}{3}$ and its minor axis subtends $90°$ at the foci. Find $(2a+2b)^2$, using the same right-triangle approach.
Show answer
$(2a+2b)^2=\dfrac{6+4\sqrt2}{9}\approx1.295$.
The minor axis subtending $90^\circ$ at a focus makes the triangle formed by the focus and the two ends of the minor axis right-angled there, so the half-angle has $\tan45^\circ=\dfrac{b}{c}=1$ — that is, $b=c$.
Then $a^2=b^2+c^2=2b^2$, so $a=b\sqrt2$, and the latus rectum $\dfrac{2b^2}{a}=b\sqrt2=\dfrac13$ gives $b=\dfrac{\sqrt2}{6}$ and $a=\dfrac13$.
Hence $2a+2b=\dfrac23+\dfrac{\sqrt2}{3}$ and squaring gives $\dfrac{(2+\sqrt2)^2}{9}=\dfrac{6+4\sqrt2}{9}$. The centre at $(2,0)$ never enters — only the shape matters.
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