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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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The common tangent to the parabola $y^2=4ax$ and $x^2=4ay$ is

$\begin{array}{1 1}(A)\;x+y+a=0 \\(B)\;x+y-a=0 \\(C)\;x-y+a=0 \\(D)\;x-y-a=0 \end{array}$

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1 Answer

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The equation of any tangent to $y^2=4ax$
$y= mx +\large\frac{a}{m}$
If it touches $x^2=4ay$ then, the equation.
$x62=4a \bigg( mx +\large\frac{a}{m}\bigg)$ has equal roots.
=> $mx^2-4am^2x-4a^2=0$
it has equal roots
=> $16a^2m^4=-16 a^2m$
=> $m=-1$
Putting $m=-1$ in $y= mx +\large\frac{a}{m}$ we get,
$y= -x-a$ or $x+y+a=0$
Hence A is the correct answer.
answered Apr 9, 2014 by meena.p
x62    may be typing mistake . Please correct it

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