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

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$y^2=4x$ 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.

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