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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

Find the equation of common tangent to the parabola $y^2=4x$ and $x^2=4y$?

$\begin{array}{1 1}(a)\;x+y=-1\\(b)\;x+2y=-2\\(c)\;y=x\\(d)\;x^2+2y=1\end{array}$

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1 Answer

$y^2=4x$ has a tangent $y=mx+\large\frac{1}{m}$
Which is also tangent to $x^2=4y$ so this tangent passes through $x^2=4y$
$\Rightarrow x^2=4(mx+\large\frac{1}{m})$
$x^2-4mx-\large\frac{4}{m}$$=0$
Roots of this equation is same as their is one tangent hence
$B^2=4AC$
$(-4m)^2=4\times 1\times -\large\frac{4}{m}$
$16m^2=-\large\frac{-16}{m}$
$m^3=-1$
Hence $m=-1$
Equation of tangent $y=-x-1$
$x+y=-1$
Hence (a) is the correct answer.
answered Feb 5, 2014 by sreemathi.v
 

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