Question

The equation of a common tangent to $y^{2}=4x$ and the curve $x^{2}+4y^{2}=8$ can be

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

Answer 1

Answer: (B)

$y^{2}=4x\&\frac{x^{2}}{8}+\frac{y^{2}}{2}=1$
Equation of tangent to above curves are respectively.
$y^{2}=mx+\frac{1}{ m}$ and $y=mx+\sqrt{8 m^{2} + 2}$
Comparing $\frac{1}{ m}=\sqrt{8 m^{2} + 2}$
$\Rightarrow m^{2}\left(8 m^{2} + 2\right)=1$
seeing the options
$m=\pm\frac{1}{2}$ satisfy the equation
$\Rightarrow y=\pm\frac{1}{2}x\pm2\Rightarrow 2y=\pm x\pm4$
i.e. $2y=x+4\&x+2y+4=0$