The equation of the common tangent drawn to t | vedclass

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(C) Let $P(t, -1/t)$ be a point on the curve $xy = -1$. The equation of the tangent at $P$ is given by $x(-1/t) + y(t) = -2$,which simplifies to $y =

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$y = x + 2$

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(C) Let $P(t, -1/t)$ be a point on the curve $xy = -1$. The equation of the tangent at $P$ is given by $x(-1/t) + y(t) = -2$,which simplifies to $y = x/t^2 + 2/t$.For this line to be a tangent to the parabola $y^2 = 8x$ (where $a = 2$),the condition $c = a/m$ must be satisfied.Here,$m = 1/t^2$ and $c = 2/t$.Substituting these into the condition: $2/t = 2 / (1/t^2) \implies 2/t = 2t^2 \implies t^3 = 1 \implies t = 1$.Substituting $t = 1$ into the tangent equation $y = x/t^2 + 2/t$,we get $y = x + 2$.

The equation of the common tangent drawn to the curves $y^2 = 8x$ and $xy = -1$ is

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