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

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

\ ty=x+at^2 where 4a=8Rightarrow 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=0Rightarrow t=0 ;or ;t=1

Hence the common tangent is y=x+2

Posted by

Deependra Verma

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