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Q.
If $y=m x+c$ is a common tangent to the parabola $y^{2}=4 \sqrt{k} x$ and the circle $2 x^{2}+2 y^{2}=k$ then the product of the slopes of such common tangents is
TS EAMCET 2020
Solution:
The equation of a tangent to $y^{2}=4 \sqrt{k} x$ is
$y=m x+\sqrt{k} / m$, where $m$ is the slope of tangent.
If it touches the $2 x^{2}+2 y^{2}=k$, 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 $y=m x+\frac{\sqrt{k}}{m}$,
the equation of common tangents are $y=x+\sqrt{k}$ and
$y=-x-\sqrt{k}$
$\therefore m_{1}=1, m_{2}=-1$
Now, $m_{1} m_{2}=1 \times-1=-1$