Question

The equation of the common tangent to $y^2=8x$ and $x^2+y^2-12x+4=0$ is

• A. $y=\pm \left(x+2\right)$
• B. $y=\pm \left(x+4\right)$
• C. $2x+3y+36=0$
• D. $3x+2y+24=0$

Answer 1

The correct option is A $y=\pm \left(x+2\right)$

$y^2=8x...............\left(1\right)$

Tangent to $\left(1\right)$ be $y=mx+\frac{2}{m}................\left(2\right)$

$=>x^2+y^2-12x+4=0$

$=>(x+6{)}^2+y^2=32...............(3)$

Tangent to $\left(3\right)$,

$y=m(x-6)\pm \left(\sqrt{3}2\right)\sqrt{1+m^2}.................\left(4\right)$

$\left(2\right)$ and $\left(4\right)$ represents the same,

$=>\frac{2}{m}=\sqrt{3}2\sqrt{1+m^2}-m\left(6\right)$

$=>{\left(\frac{{6m}^2+2}{m}\right)}^2=32(1+m^2)$

$=>9m^4+1+6m^2=8m^2+8m^4$

$=>m^4-2m^2+1=0$

$=>m^2=\frac{2\pm \sqrt{4-4}}{2}$

$=>m^2=1$

Tangent which are common o $\left(1\right)$ and $\left(3\right)$ are,

$=>y=\pm x\pm 2$

$=>y=\pm (x+2).$