Question

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

• A. $3y=9x+2$
• B. $y=2x+1$
• C. $2y=x+8$
• D. $y=x+2$

Answer 1

Answer: (D)

$y=x+2$

Equation of a tangent to parabola $y^2=8x$ is

$y=mx+\frac{2}{m}\Rightarrow \left(1\right)$

Now solving $\left(1\right)$ with $xy=-1$

$x(mx+\frac{2}{m})=-1$

$\Rightarrow m{x}^2+\left(\frac{2}{m}\right)x+1=0$

Now for $\left(1\right)$ to be tangent to $xy=-1$, discriminant of above quadratic should be zero.

$\Rightarrow (\frac{2}{m}{)}^2-4m=0$

$\Rightarrow m^3=1\Rightarrow m=1$, only real solution

Hence required common tangent is $y=x+2$