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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?
$\left( {4,\,\, - \,\,\sqrt 6 } \right)$
$\left( {\sqrt 6 ,\,\,1} \right)$
$\left( {\frac{1}{2},\,\,\frac{1}{{\sqrt 6 }}} \right)$
$\left( {\frac{\pi }{6}\,\,\pi } \right)$
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 $(4,-\sqrt{6})$