The equation of the common tangent drawn to the curves $y^2=8 x$ and $x y=-1$ is
The equation of the common tangent drawn to the curves $y^2=8 x$ and $x y=-1$ is
$y = 2x + 1$
$2y = x + 6$
$y = x + 2$
$3y = 8x + 2$
Solution
Let $P\left(t, \frac{-1}{t}\right)$ be a point on $x y=-1$ Equation of tangent of $P$
$\begin{aligned}
& =\frac{1}{2}\left[x\left(\frac{-1}{t}\right)+x y\right]=-1-x+t^2 y=-2 t \\
& y=\frac{x}{t^2}+\frac{2}{t}
\end{aligned}$
This is also at tangent of $y^2=8 x$
$\begin{aligned}
& \text { i.e. } c=\frac{a}{m} \\
& \frac{2}{t}=\frac{2}{\frac{1}{t^2}} \\
& \Rightarrow t^3=1 \\
& \Rightarrow t=1
\end{aligned}$
Hence, the equation of common tangent
$\begin{aligned}
& y=\frac{x}{1}+\frac{2}{1} \\
& y=x+2
\end{aligned}$