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The equation of common tangent to the curves $y^2=8x$ and $xy=-1$ is
$\begin{array}{1 1}(A)\;3y=9x+2 \\(B)\;y=2x+1 \\(C)\;2y=x+8 \\(D)\;y=x+2 \end{array}$
jeemain
math
class11
coordinate-geometry-conic -sections
parabola
ch11
medium
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asked
Apr 10, 2014
by
meena.p
retagged
Sep 27, 2014
by
sharmaaparna1
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1 Answer
Any tangent to $y^2=8x$ is
$Y=mx+2/m$
it should also be tangent two $xy=-1$
$\bigg(mx+\large\frac{2}{m}\bigg)$$x=-1$
$m^2x^2+2x+m=0$
$D=0 => 4-4(+m)(m^2)=0$
$\qquad= m^3=+1$
=>$m=1$
So, the equation is $y=x+2$
Hence D is the correct answer.
answered
Apr 10, 2014
by
meena.p
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