The equation of the common tangent to the&nbs | vedclass

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Question

Any point on $y^2 = 8x$ is $(2t_2 , 4t)$ where the tangent is $yt = x + 2t_2$.Solving it with $xy = -1, y(yt - 2t_2 ) = -1$ or $ty_2 - 2t_2y +1 = 0$ .

Vedclass · Accepted Answer

$y = x +2$

Vedclass · Answer

Any point on $y^2 = 8x$ is $(2t_2 , 4t)$ where the tangent is $yt = x + 2t_2$.Solving it with $xy = -1, y(yt - 2t_2 ) = -1$ or $ty_2 - 2t_2y +1 = 0$ . For common tangent, it should have equal roots.

$	herefore 4t^4 - 4t = 0 \Rightarrow t = 0, 1$ .

$	herefore$ The common tangent is $y = x +2$ , (whent $t= 0$ , it is $x = 0$ which can touch $xy = -1$ atinfinity only).

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

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