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10-2. Parabola, Ellipse, Hyperbola
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રેખા ${\text{2x}}\,\, + \;\,\sqrt {\text{6}} y\,\, = \,\,2$ એ વક્ર $\,{x^2}\, - \,\,2{y^2}\,\, = \,\,4\,\,$ ને કયા બિંદુ આગળ સ્પર્શે  છે?

A

$\left( {4,\,\, - \,\,\sqrt 6 } \right)$

B

$\left( {\sqrt 6 ,\,\,1} \right)$

C

$\left( {\frac{1}{2},\,\,\frac{1}{{\sqrt 6 }}} \right)$

D

$\left( {\frac{\pi }{6}\,\,\pi } \right)$

Solution

Given the line $2 \pi+\sqrt{ } 6 y =2$ touches the hyperbola $x ^2-2 y ^2=4$.

We have to find the point of contact.

Conaider $2 x+\sqrt{6 y}=2$

$=2 x =2-\sqrt{6 y}$

$=\pi=\frac{2-\sqrt{6 y}}{2}$

Substitute $in x ^1-2 y ^1=4 we$ get

$\frac{\left(2-\sqrt{6 y)^2}\right.}{4}-2 y^2=4$

$=\frac{4+6 y^2-4 \sqrt{6 y}}{4}-2 y^2=4$

$=\frac{4+6 y^2-4 \sqrt{6 y}-8 y^2}{4}=4$

$=4-2 y^2-4 \sqrt{6 y}=16$

$=2 y ^2+4 \sqrt{6}+12=0$

$=y^2+2 \sqrt{6} y+6=0$

$=( y +\sqrt{6})^2=0$

$=y+\sqrt{6}=0$

$=y=-\sqrt{6}$

Substitutey in $W$ we get

$\pi=\frac{2-\sqrt{6} \times(-\sqrt{6})}{2}$

$=\frac{2+6}{2}$

$=\frac{8}{2}$

$\therefore x=4$

Thus the point of contact is $(4,-\sqrt{6})$

Standard 11
Mathematics

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