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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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The equation of common tangent to the curves $y^2=8x$ and $xy=-1$ is

$\begin{array}{1 1}(A)\;3y=9x+2 \\(B)\;y=2x+1 \\(C)\;2y=x+8 \\(D)\;y=x+2 \end{array}$

Can you answer this question?
 
 

1 Answer

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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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