The equation of the common tangent to the cur | vedclass

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Question

(D) Let the equation of the tangent to the parabola $y^2 = 8x$ be $y = mx + \frac{2}{m}$,where $m 
eq 0$.This line is also a tangent to the hyperbola

Vedclass · Accepted Answer

$y = x + 2$

Vedclass · Answer

(D) Let the equation of the tangent to the parabola $y^2 = 8x$ be $y = mx + \frac{2}{m}$,where $m 
eq 0$.This line is also a tangent to the hyperbola $xy = -1$,which can be written as $y = -\frac{1}{x}$.Substituting $y = mx + \frac{2}{m}$ into $xy = -1$,we get $x(mx + \frac{2}{m}) = -1$,which simplifies to $mx^2 + \frac{2}{m}x + 1 = 0$.For the line to be a tangent,the discriminant of this quadratic equation must be zero:$D = (\frac{2}{m})^2 - 4(m)(1) = 0$$\frac{4}{m^2} - 4m = 0$$4 = 4m^3$ $\Rightarrow m^3 = 1$ $\Rightarrow m = 1$.Substituting $m = 1$ into the tangent equation $y = mx + \frac{2}{m}$,we get $y = 1x + \frac{2}{1}$,which is $y = x + 2$.

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

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