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 $y=mx+\sqrt{8m^2+2}$
Comparing $\frac{1}{m}=\sqrt{8m^2+2}$
$\Rightarrow m^2\left(8m^2+2\right)=1$
seeing the options
$m=\pm \frac{1}{2}$ satisfy the equation
$\Rightarrow y=\pm \frac{1}{2}x\pm 2\Rightarrow 2y=\pm x\pm 4$
i.e. $2y=x+4\&x+2y+4=0$