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Hyperbola

Section: Mathematics

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Question : 60 of 98

Marks: +1, -0

The point

P(-2\sqrt{6},\sqrt{3})

lies on the hyperbola

\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1

having eccentricity

\frac{\sqrt{5}}{2}

. If the tangent and normal at P to the hyperbola intersect its conjugate axis at the point Q and R respectively, then Q R is equalto

[26 Aug 2021 Shift 2]

$4\sqrt{3}$
6
$6\sqrt{3}$
$3\sqrt{6}$

Solution:

a^{2}+b^{2}=a^{2}e^{2}

a^{2}+b^{2}=a^{2}\left(\frac{5}{4}\right)

4b^{2}=a^{2}

\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1

∵ Point

P(-2\sqrt{6},\sqrt{3})

lies on given ellips.

\therefore \frac{24}{4b^{2}}-\frac{3}{b^{2}}=1

\Rightarrow b^{2}=3

\Rightarrow a^{2}=12

Equation of tangent at P

\frac{x(-2\sqrt{6})}{12}-\frac{y(\sqrt{3})}{3}=1

For conjugate axis, put x = 0

∴

Q(0,-\sqrt{3})

Equation of normal at P

\frac{y-\sqrt{3}}{x+2\sqrt{6}}=\left(\frac{-1}{\sqrt{3}}\right)\times\left(\frac{12}{-2\sqrt{6}}\right)

⇒

y-\sqrt{3}=2\sqrt{12}=4\sqrt{3}

\Rightarrow y=5\sqrt{3}

\therefore R(0,5\sqrt{3})

\Rightarrow QR=6\sqrt{3}

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