Question

If the line $y=mx+c$ touches $x^2-y^2=1$ and $y^2=4x,$ then $m^2$ is equal to

• A. $\frac{\sqrt{5}-1}{2}$
• B. $\frac{1+\sqrt{5}}{2}$
• C. $5$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $\frac{1+\sqrt{5}}{2}$

Tangent of slope $m$ to the parabola $y^2=4x$ is
$y=mx+\frac{1}{m}\dots \left(1\right)$

Tangent of slope $m$ to the hyperbola $x^2-y^2=1$ is
$y=mx\pm \sqrt{m^2-1}\dots \left(2\right)$

As lines $\left(1\right)$ and $\left(2\right)$ are identical, then
$\frac{1}{m^2}=m^2-1\Rightarrow m^4-m^2-1=0\Rightarrow m^2=\frac{1\pm \sqrt{5}}{2}\therefore m^2=\frac{1+\sqrt{5}}{2}(\because m^2>0)$