Question

Equation of one of the common tangent of $y^2=4x$ and $\frac{x^2}{4}+\frac{y^2}{3}=1$ is

• A. $x+2y+4=0$
• B. $x+2y-4=0$
• C. $x-2y-4=0$
• D. $2x+y-4=0$

Answer 1

Answer: (A)

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