A double ordinate PQ of the hyperbola (x2/a2)-(y2/b2)=1 is such that triangle O P Q is equilateral, O being the centre of the hyperbola. Then the eccentricity e satisfies the relation

Question

A double ordinate PQ of the hyperbola (x2/a2)-(y2/b2)=1 is such that triangle O P Q is equilateral, O being the centre of the hyperbola. Then the eccentricity e satisfies the relation (A) 1<e<(2/√3) (B) e=(2/√3) (C) e=(√3/2) (D) e>(2/√3)

Tardigrade Admin · Accepted Answer

Let, P =( a sec θ, b tan θ) angle POM = tan 30°=(1/√3)=(b/a) sin θ ⇒ sin θ (a/b √3) ⇒ (b2/a2)>(1/3) ⇒ 1+(b2/a2)>(4/3) ⇒ e>(2/√3)

Q. A double ordinate $PQ$ of the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ is such that $△ OPQ$ is equilateral, $O$ being the centre of the hyperbola. Then the eccentricity e satisfies the relation

$1 < e < \frac{2}{3}$

$e = \frac{2}{3}$

$e = \frac{3}{2}$

$e > \frac{2}{3}$

Let, $P = (a sec θ, b tan θ) ∠ POM = tan 3 0^{\circ} = \frac{1}{3} = \frac{b}{a} sin θ$
$\Rightarrow sin θ \frac{a}{b 3}$
$\Rightarrow \frac{b ^{2}}{a ^{2}} > \frac{1}{3}$
$\Rightarrow 1 + \frac{b ^{2}}{a ^{2}} > \frac{4}{3}$
$\Rightarrow e > \frac{2}{3}$

Q. A double ordinate PQ of the hyperbola a2x2​−b2y2​=1 is such that △OPQ is equilateral, O being the centre of the hyperbola. Then the eccentricity e satisfies the relation

1<e<3​2​

e=3​2​

e=23​​

e>3​2​