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

Question

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

Tardigrade Admin · Accepted Answer

tan 30°=(b tan θ/a sec θ) ⇒ (b/a)=(1/ sin θ √3) e=√1+(1/3 sin 2 θ)>√1+(1/3)

⇒ e>(2/√3)(0< sin 2 θ<1)

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

$1 < e < 2/ 3$

$e = 2/ 3$

$e = 23$

$e > 2/ 3$

$tan 3 0^{\circ} = \frac{b t a n θ}{a s e c θ}$
$\Rightarrow \frac{b}{a} = \frac{1}{s i n θ 3}$
$e = 1 + \frac{1}{3 s i n ^{2} θ} > 1 + \frac{1}{3}$

$\Rightarrow e > \frac{2}{3} (0 < sin^{2} θ < 1)$

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

1<e<2/3​

e=2/3​

e=23​

e>2/3​