The common tangents to the circle x^2+y^2=2 a | vedclass

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(D) The equation of a tangent to the parabola $y^2=8x$ is $y=mx+\frac{2}{m}$.For this to be a tangent to the circle $x^2+y^2=2$,the perpendicular dist

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$15$

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(D) The equation of a tangent to the parabola $y^2=8x$ is $y=mx+\frac{2}{m}$.For this to be a tangent to the circle $x^2+y^2=2$,the perpendicular distance from the center $(0,0)$ to the line $mx-y+\frac{2}{m}=0$ must be equal to the radius $\sqrt{2}$.$\left|\frac{2/m}{\sqrt{m^2+1}}ight|=\sqrt{2}$ $\Rightarrow \frac{4}{m^2(m^2+1)}=2$ $\Rightarrow m^4+m^2-2=0$.Let $m^2=t$,then $t^2+t-2=0 \Rightarrow (t+2)(t-1)=0$. Since $t=m^2 > 0$,we have $m^2=1$,so $m=\pm 1$.The common tangents are $y=x+2$ and $y=-x-2$.The points of contact $P, Q$ on the circle $x^2+y^2=2$ are found by the chord of contact formula $xx_1+yy_1=r^2$. For the tangent $x-y+2=0$,the point of contact is $(-1, 1)$. For $x+y+2=0$,it is $(-1, -1)$. Thus $P=(-1, 1)$ and $Q=(-1, -1)$.The points of contact $R, S$ on the parabola $y^2=8x$ are found using $yy_1=4(x+x_1)$. For $y=x+2$,$y_1y=4(x+x_1)$ $\Rightarrow 1(y)=4(x+1)$ $\Rightarrow y=4x+4$. Comparing with $y=x+2$,we get $x=2, y=4$. For $y=-x-2$,we get $x=2, y=-4$. Thus $R=(2, 4)$ and $S=(2, -4)$.The quadrilateral $PQRS$ is a trapezium with parallel sides $PQ$ and $RS$.Length $PQ = |1 - (-1)| = 2$.Length $RS = |4 - (-4)| = 8$.Height $h = |2 - (-1)| = 3$.Area $= \frac{1}{2}(PQ+RS) 	imes h = \frac{1}{2}(2+8) 	imes 3 = \frac{1}{2}(10) 	imes 3 = 15$.

The common tangents to the circle $x^2+y^2=2$ and the parabola $y^2=8x$ touch the circle at the points $P, Q$ and the parabola at the points $R, S$. Then the area of the quadrilateral $PQRS$ is

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