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Q.
The equations of the common tangent to the ellipse, $x^2+4 y^2=8 \&$ the parabola $y^2=4 x$
Conic Sections
Solution:
$\frac{x^2}{8}+\frac{y^2}{2}=1 $......(i)
$y^2=4 x$......(ii)
let $m$ be the slope of common tangent equation of tangent of slope $m$ to parabola $y^2=4 x$......(iii)
$\therefore y = mx +\frac{1}{ m }$
since this is also tangent to ellipse
$\left(\frac{1}{m}\right)^2=8 m^2+2$
$\Rightarrow 1=8 m^4+2 m^2-1=0 $
$\Rightarrow 8 m^4+2 m^2-1=0$
$m^4+\frac{1}{4} m^2-\frac{1}{8}=0$
$\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 2 y-x=4 $
$y=-\frac{x}{2}-2 \therefore 2 y+x+4=0$