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Q.
Equation of one of the common tangent of $y^2=4 x$ and $\frac{x^2}{4}+\frac{y^2}{3}=1$ is
Conic Sections
Solution:
Tangent to $y^2=4 x$ is $y=m x+\frac{1}{m}$ .....(i)
and tangent to $\frac{x^2}{4}+\frac{y^2}{3}=1$ will be $y=m x \pm \sqrt{4 m^2+3}$....(ii)
$\Theta$ (i) and (ii) represent same line for common tangent
$ \frac{1}{m}=\sqrt{4 m^2+3} $
$\Rightarrow 1=4 m^4+3 m^2 \Rightarrow 4 m^4+3 m^2-1=0$
$\Rightarrow \left(4 m^2-1\right)\left(m^2+1\right)=0 \Rightarrow m^2=\frac{1}{4} \Rightarrow m= \pm \frac{1}{2}$
$\therefore$ Common tangent will be :
$y=\frac{1}{2} x+2 \Rightarrow x-2 y+4=0 $
$y=\frac{-1}{2} x-2 \Rightarrow x+2 y+4=0$