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Hyperbola question

2024 · 1 Feb · Shift 1 · Q43

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Hyperbola
2024 · 1 Feb · Shift 1 · Q43

Hyperbola question

2024 · 1 Feb · Shift 1 · Q43

JEE MainMathematicsHyperbolaMCQ+4 / −1

For

0\lt \theta\lt \pi / 2

, if the eccentricity of the hyperbola

x^2-y^2 \operatorname{cosec}^2 \theta=5

is

\sqrt{7}

times eccentricity of the ellipse

x^2 \operatorname{cosec}^2 \theta+y^2=5

, then the value of

\theta

is :

A
$\frac{\pi}{6}$
B
$\frac{5 \pi}{12}$
C
$\frac{\pi}{3}$
D
$\frac{\pi}{4}$

View written solutionFree

Correct answer: C

Write both conics in standard form

Given hyperbola: $x^2-y^2\csc^2\theta=5$ Divide by $5$ :

\implies \frac{x^2}{5}-\frac{y^2}{5\sin^2\theta}=1$$ So for the hyperbola, $$a_h^2=5,\qquad b_h^2=5\sin^2\theta$$ Hence its eccentricity is $$e_h=\sqrt{1+\frac{b_h^2}{a_h^2}}=\sqrt{1+\sin^2\theta}$$ Now the ellipse: $$x^2\csc^2\theta+y^2=5$$ Divide by $5$: $$\frac{x^2\csc^2\theta}{5}+\frac{y^2}{5}=1 \implies \frac{x^2}{5\sin^2\theta}+\frac{y^2}{5}=1$$ So for the ellipse, $$a_e^2=5,\qquad b_e^2=5\sin^2\theta$$ since $0<\theta<\frac\pi2$, we have $\sin^2\theta<1$, so $5>5\sin^2\theta$ and major axis is along $y$. Thus eccentricity of ellipse is $$e_e=\sqrt{1-\frac{b_e^2}{a_e^2}}=\sqrt{1-\sin^2\theta}=\cos\theta$$ 2. **Use the given relation** Given: $$e_h=\sqrt7\,e_e$$ So, $$\sqrt{1+\sin^2\theta}=\sqrt7\cos\theta$$ Squaring both sides, $$1+\sin^2\theta=7\cos^2\theta$$ Using $\cos^2\theta=1-\sin^2\theta$, $$1+\sin^2\theta=7(1-\sin^2\theta)$$ $$1+\sin^2\theta=7-7\sin^2\theta$$ $$8\sin^2\theta=6$$ $$\sin^2\theta=\frac34$$ Since $0<\theta<\frac\pi2$, $$\sin\theta=\frac{\sqrt3}{2}$$ Hence, $$\theta=\frac\pi3$$ 3. **Check options** - A: $\frac\pi6$ ❌ - B: $\frac{5\pi}{12}$ ❌ - C: $\frac\pi3$ ✅ - D: $\frac\pi4$ ❌ Therefore, the correct option is **C**.

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