JEE Main 2016 (Offline) — Previous Year Question.
Problem
The eccentricity of the hyperbola whose latus rectum is 8 and whose conjugate axis is equal to half of the distance between the foci is:
Key insight. Both given facts — the latus rectum and the conjugate-axis relation — translate into two equations relating $a$, $b$, and $c$. Combined with the hyperbola’s own identity $c^2=a^2+b^2$, that’s enough equations to eliminate everything except $b$, letting eccentricity fall out at the end.
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Approach
For a hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, translate “latus rectum is 8” into $\dfrac{2b^2}a=8$, and “conjugate axis equals half the distance between the foci” into $2b=\dfrac12(2c)$. These give $b^2$ in terms of $a$, and $c$ in terms of $b$. Substituting both into the identity $c^2=a^2+b^2$ eliminates $a$ and $c$ entirely, leaving a single equation in $b$ alone — solve it, then compute $e=c/a$.
Solution
Step 1 — Translate the latus rectum condition
$$\text{Latus rectum} = \frac{2b^2}a = 8 \implies b^2=4a \implies a=\frac{b^2}4$$
Step 2 — Translate the conjugate-axis condition
The conjugate axis has length $2b$, and the distance between the foci is $2c$:
$$2b = \frac12(2c) = c \implies c=2b$$
Step 3 — Use c² = a² + b² to eliminate a and c
$$c^2=a^2+b^2 \implies (2b)^2 = \left(\frac{b^2}4\right)^2+b^2$$
$$4b^2 = \frac{b^4}{16}+b^2$$
Step 4 — Solve for b
$$3b^2 = \frac{b^4}{16}$$
Since $b\ne0$, divide both sides by $b^2$:
$$3 = \frac{b^2}{16} \implies b^2=48 \implies b=4\sqrt3$$
Step 5 — Find a and c
$$a=\frac{b^2}4=\frac{48}4=12, \qquad c=2b=8\sqrt3$$
Step 6 — Compute the eccentricity
$$e=\frac ca=\frac{8\sqrt3}{12}=\frac{2\sqrt3}3=\frac2{\sqrt3}$$
Answer
$$\frac2{\sqrt3}$$
Common mistakes
- Mixing up the conjugate axis with the transverse axis. The conjugate axis has length $2b$, not $2a$ — using $a$ in Step 2 instead of $b$ leads to an entirely different (and inconsistent) system of equations.
- Forgetting to divide by $b^2$ before it’s safe to do so. Step 4’s equation $3b^2=\dfrac{b^4}{16}$ has $b=0$ as an extraneous root alongside $b^2=48$; dividing through by $b^2$ is valid precisely because $b\ne0$ for a genuine hyperbola.
- Computing e as a/c instead of c/a. Eccentricity is always $e=c/a$ for a hyperbola (with $e>1$) — inverting the ratio gives a fraction less than 1, which isn’t possible for a hyperbola and should be a red flag that something’s been flipped.
Practise next
- Find the eccentricity of a hyperbola whose latus rectum is 8, but whose conjugate axis is one-third of the distance between the foci.
Show answer
$\dfrac{3\sqrt2}4$. Now $2b=\dfrac13(2c)\Rightarrow c=3b$, so $c^2=a^2+b^2$ gives $9b^2=a^2+b^2\Rightarrow a^2=8b^2$.
With $b^2=4a$ still holding (same latus rectum), $a^2=8(4a)=32a\Rightarrow a=32$, so $b^2=128$ and $c=3\sqrt{128}=24\sqrt2$.
$e=\dfrac ca=\dfrac{24\sqrt2}{32}=\dfrac{3\sqrt2}4$.
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