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Question

(A) For the parabola $y^2 = 8x$,the equation of the tangent at point $(2t^2, 4t)$ is $yt = x + 2t^2$.This line is also a tangent to the hyperbola $xy

Vedclass · Accepted Answer

$y = x + 2$

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(A) For the parabola $y^2 = 8x$,the equation of the tangent at point $(2t^2, 4t)$ is $yt = x + 2t^2$.This line is also a tangent to the hyperbola $xy = -1$. Substituting $x = \frac{yt - 2t^2}{1}$ into $xy = -1$,we get $y(\frac{yt - 2t^2}{1}) = -1$,which simplifies to $ty^2 - 2t^2y + 1 = 0$.For the line to be a tangent,the discriminant of this quadratic equation must be zero:$D = (-2t^2)^2 - 4(t)(1) = 0$$4t^4 - 4t = 0$$4t(t^3 - 1) = 0$Since $t 
eq 0$ (as $t=0$ gives $x=0$,which is not a tangent to $xy=-1$),we have $t^3 = 1$,so $t = 1$.Substituting $t = 1$ into the tangent equation $yt = x + 2t^2$,we get $y(1) = x + 2(1)^2$,which simplifies to $y = x + 2$.

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

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