Question

The equations of the common tangent to the ellipse, $x^2+4y^2=8\&$ the parabola $y^2=4x$

• A. $2y-x=8$
• B. $2y+x=4$
• C. $2y+x+4=0$
• D. $2y+x=0$

Answer 1

Answer: (C)

$\frac{x^2}{8}+\frac{y^2}{2}=1$......(i)
$y^2=4x$......(ii)
let $m$ be the slope of common tangent equation of tangent of slope $m$ to parabola $y^2=4x$......(iii)
$\therefore y=mx+\frac{1}{m}$
since this is also tangent to ellipse
${\left(\frac{1}{m}\right)}^2=8m^2+2$
$\Rightarrow 1=8m^4+2m^2-1=0$
$\Rightarrow 8m^4+2m^2-1=0$
$m^4+\frac{1}{4}{m}^2-\frac{1}{8}=0$
$\left(m^2+\frac{1}{2}\right)\left(m^2-\frac{1}{4}\right)=0$
$\because m^2=\frac{1}{4}$
$\therefore m=\pm \frac{1}{2}$
$y=\frac{x}{2}+2\therefore 2y-x=4$
$y=-\frac{x}{2}-2\therefore 2y+x+4=0$