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

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

Satisfies

A

1 < e <23\frac{2}{\sqrt{3}}

B

e=23e = \frac{2}{\sqrt{3}}

C

e=32e = \frac{\sqrt{3}}{2}

D

e>23e > \frac{2}{\sqrt{3}}

Answer

e>23e > \frac{2}{\sqrt{3}}

Explanation

Solution

∵\because PQ is the doubled ordinate. Let MP=MQ=lMP = MQ = lgiven that

ΔOPQ\Delta OPQs an equilateral then OP = OQ = PQ

⇒ (OP)2=(OQ)2=(PQ)2(OP)^{2} = (OQ)^{2} = (PQ)^{2}

⇒ a2b2(b2+l2)+l2=a2b2(b2+l2)+l2=4l2\frac{a^{2}}{b^{2}}\left( b^{2} + l^{2} \right) + l^{2} = \frac{a^{2}}{b^{2}}\left( b^{2} + l^{2} \right) + l^{2} = 4l^{2}

⇒a2b2(b2+l2)=3l2\frac{a^{2}}{b^{2}}\left( b^{2} + l^{2} \right) = 3l^{2}

⇒a2=l2(3−a2b2)a^{2} = l^{2}\left( 3 - \frac{a^{2}}{b^{2}} \right)

⇒l2=a2b2(3b2−a2)>0l^{2} = \frac{a^{2}b^{2}}{\left( 3b^{2} - a^{2} \right)} > 0∴ 3b2−a2>03b^{2} - a^{2} > 0

⇒ 3b2>a23b^{2} > a^{2}

⇒ 3a2(e2−1)>a23a^{2}\left( e^{2} - 1 \right) > a^{2}

⇒ e2>43e^{2} > \frac{4}{3} ∴e>23\therefore e > \frac{2}{\sqrt{3}}