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

3 y = 9 x + 2

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y = 2 x + 1

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2 y = x + 8

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y = x + 2

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Solution

The correct option is D $y = x + 2$
Equation of tangent to $y^{2} = 8 x$ is $y = m x + \frac{2}{m}$
To be tangent to the curve $x y = - 1$
$x (m x + \frac{2}{m}) = - 1$
For equal roots the discriminant of quadratic equation has to be zero (condition of equal roots)
$m^{2} x^{2} + 2 x + m = 0$
$4 - 4 m^{3} = 0$
$m = 1$
option D

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Similar questions

y^{2} = 8 x and x y = - 1

y^{2} = 8 x a n d x y = - 1

x y = - 1

y^{2} = 8 x

The equation of the common tangent to the curves is $y^{2} = 8 x a n d x y = - 1$ is

y^{2} = 16 x

x y = - 4

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