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Q.
The equation of the common tangent to the curves $y^{2}=$ $8 x$ and $x y=-1$ is
Conic Sections
Solution:
Any tangent to the parabola $y^{2}=8 x$ is
$y=m x+\frac{2}{m}$.
Clearly (a), (b) and (d) satisfy the equation (by $m=$ $3,2,1)$.
For $x y=-1$
Equation of tangent with slope $m$, to $x y=-1$ are
$y-\sqrt{m}=m\left(x+\frac{1}{\sqrt{m}}\right)$
$\left(\because \frac{d y}{d x}=\frac{1}{x^{2}}=m \therefore x=\pm \frac{1}{\sqrt{m}}, y=\mp \sqrt{m}\right)$
Clearly (d) satisfies these equations.