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Question: If PQ is a double ordinate of hyperbola \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\) such that ...

If PQ is a double ordinate of hyperbola x2a2−y2b2=1\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 such that CPQ is an equilateral triangle, C being the centre of the hyperbola. Then the eccentricity e of the hyperbola satisfies

A

1<e<2/31 < e < 2/\sqrt{3}

B

e=2/3e = 2/\sqrt{3}

C

e=3/2e = \sqrt{3}/2

D

e>2/3e > 2/\sqrt{3}

Answer

e>2/3e > 2/\sqrt{3}

Explanation

Solution

Let P(a⥂sec⁡θ,btan⁡θ)P(a ⥂ \sec\theta,b\tan\theta); Q(asec⁡θ,−btan⁡θ)Q(a\sec\theta, - b\tan\theta) be end points of double ordinates and C(0,0)C(0,0) is the centre of the hyperbola

Now PQ=2btan⁡θPQ = 2b\tan\theta; CQ=CP=a2sec⁡2θ+b2tan⁡2θCQ = CP = \sqrt{a^{2}\sec^{2}\theta + b^{2}\tan^{2}\theta}

Since CQ=CP=PQCQ = CP = PQ, ∴\therefore 4b2tan⁡2θ=a2sec⁡2θ+b2tan⁡2θ4b^{2}\tan^{2}\theta = a^{2}\sec^{2}\theta + b^{2}\tan^{2}\theta

⇒ 3b2tan⁡2θ=a2sec⁡2θ3b^{2}\tan^{2}\theta = a^{2}\sec^{2}\theta ⇒ x2a2−y2b2=1a2−1b2\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = \frac{1}{a^{2}} - \frac{1}{b^{2}}

⇒ 3a2(e2−1)sin⁡2θ=a23a^{2}(e^{2} - 1)\sin^{2}\theta = a^{2} ⇒ 3(e2−1)sin⁡2θ=13(e^{2} - 1)\sin^{2}\theta = 1

⇒ 13(e2−1)=sin⁡2θ<1\frac{1}{3(e^{2} - 1)} = \sin^{2}\theta < 1 (∵sin⁡2θ<1)(\because\sin^{2}\theta < 1)

⇒ 1e2−1<3\frac{1}{e^{2} - 1} < 3 ⇒ x2a4−y2b4=1a2+1b2\frac{x^{2}}{a^{4}} - \frac{y^{2}}{b^{4}} = \frac{1}{a^{2}} + \frac{1}{b^{2}} ⇒ e2>43e^{2} > \frac{4}{3} ⇒ e>23e > \frac{2}{\sqrt{3}}