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10-2. Parabola, Ellipse, Hyperbola
hard
અતિવલય $4{x^2} - {y^2} = 36$ ને બિંદુ $P$ અને $Q$ આગળ સ્પર્શકો દોરવામાં આવે છે. જો આ સ્પર્શકો બિંદુ $T\left( {0,3} \right)$ આગળ છેદે તો $\Delta PTQ$ નું ક્ષેત્રફળ . . . . . .છે. .
A
$54\sqrt 3 $
B
$60\sqrt 3 $
C
$36\sqrt 5 $
D
$45$$\sqrt 5 $
(JEE MAIN-2018)
Solution
(4) Here equation of hyperbola is $\frac{{{x^2}}}{9} – \frac{{{y^2}}}{{36}} = 1$
Now, $PQ$ is the chord of contant
$\therefore $ Equation of $PQ$ is $\,:\frac{{x\left( 0 \right)}}{9} – \frac{{y\left( 3 \right)}}{{36}} = 1$
$ \Rightarrow y = – 12$
$\therefore $ Area of $\Delta PQT = \frac{1}{2} \times TR \times PQ$
$\because $ $P \equiv \left( {3\sqrt 5 , – 12} \right)\,\,\,\,\,\,\therefore TR = 3 + 12 = 15$
$\therefore $ Area of $\Delta PQT = \frac{1}{2} \times 15 \times 6\sqrt 5 = 45\sqrt 5 $ sq.units
Standard 11
Mathematics
