# Equation of common tengent to the curve y^2=8x and xy=-1 is

Solution: Equation of a tangent at$\\(at^2,2at)$ to $y^2=8x$ is

$\\ ty=x+at^2$ where $4a=8\Rightarrow a=2$

$\\ ty=x+2t^2$ which intesect the curve $xy=-1$ at the point given by$\\ x(x+2t^2)/t=-1$

$\\x^2+2t^2x+t=0$ and will be a tangent to the curve if the roots of this quadratic equation are equal , for which $4t^4-4t=0\Rightarrow t=0 ;or ;t=1$

Hence the common tangent is $y=x+2$

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