A double ordinate PQ of the hyperbola ^ 2 ^ 2 - ^ 2 ~b^ 2 =1 is such that Δ OPQ is equilateral, O being the centre of the hyperbola. Then the eccentricity e satisfies the relation

Correct answer: D. e> 2 3 Hint: P (a θ, b θ) POM= 30^ ^ ^ = 1 3 = b a θ θ= a b 3 b^ 2 a^ 2 >(1)/(3) 1+ b^ 2 a^ 2 >(4)/(3) e> 2 3

JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Hyperbola

JEE Main · Mathematics · Hyperbola

A double ordinate PQ of the hyperbola ^ 2 ^ 2 - ^ 2 ~b^ 2 =1 is...

A double ordinate

\mathrm{PQ}

of the hyperbola

\frac{\mathrmx^{2}}{\mathrma^{2}}-\frac{\mathrmy^{2}}{\mathrm{~b}^{2}}=1

is such that

\Delta \mathrm{OPQ}

is equilateral,

\mathrm{O}

being the centre of the hyperbola. Then the eccentricity e satisfies the relation A double ordinate PQ of the hyperbola ^ 2 ^ 2 - ^ 2 ~b^ 2 =1 is such that Δ OPQ is equilateral, O being the centre of

A double ordinate PQ of the hyperbola ^ 2 ^ 2 - ^ 2 ~b^ 2 =1 is such that Δ OPQ is equilateral, O being the centre of

Options

A $1 < \mathrm{e} < \frac{2}{\sqrt{3}}$
B $\mathrm{e}=\frac{2}{\sqrt{3}}$
C $\mathrm{e}=\frac{\sqrt{3}}{2}$
D $\mathrm{e}>\frac{2}{\sqrt{3}}$

Correct answer

D. $\mathrm{e}>\frac{2}{\sqrt{3}}$

Step-by-step solution

Hint:

P \equiv(a \sec \theta, b \tan \theta) \angle \mathrm{POM}=\tan 30^{{}^{{}^{\circ}}}=\frac{1}{\sqrt{3}}=\frac{\mathrm{b}}{\mathrm{a}} \sin \theta\Rightarrow \sin \theta=\frac{a}{b \sqrt{3}}\Rightarrow \frac{b^{2}}{a^{2}}>\frac{1}{3} \quad \Rightarrow 1+\frac{b^{2}}{a^{2}}>\frac{4}{3} \quad \Rightarrow e>\frac{2}{\sqrt{3}}

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