Question

If $y=mx+c$ is a common tangent to the parabola $y^2=4\sqrt{k}x$ and the circle $2x^2+2y^2=k$ then the product of the slopes of such common tangents is

• A. $-2$
• B. $\frac{k+2}{3}$
• C. $-1$
• D. $\frac{k}{2}$

Answer 1

Answer: (C)

The equation of a tangent to $y^2=4\sqrt{k}x$ is
$y=mx+\sqrt{k}/m$, where $m$ is the slope of tangent.
If it touches the $2x^2+2y^2=k$, then
$\left|\frac{\sqrt{k}/m}{\sqrt{1+m^2}}\right|=\sqrt{\frac{k}{2}}$
$\Rightarrow m\sqrt{1+m^2}=\sqrt{2}$
$\Rightarrow m^4+m^2-2=0$
$\Rightarrow \left(m^2+2\right)\left(m^2-1\right)=0$
$\Rightarrow m=\pm 1$
Substituting these values in $y=mx+\frac{\sqrt{k}}{m}$,
the equation of common tangents are $y=x+\sqrt{k}$ and
$y=-x-\sqrt{k}$
$\therefore m_1=1,m_2=-1$
Now, $m_1{m}_2=1\times -1=-1$