Discussion Forum

 The equation of one of the common tangent to the parabola y² =8x   and x²+y² -12x+4=0  is 

Answer:

Explanation:

Any tangent to parabola y² =8x is  $y= mx+\frac{2}{m}$   .............(i)

 It touches the circle  $x^{2}+y^{2}-12x+4=0$

 if the length of perpendicular from the centre (6,0) is equal to radis  $\sqrt{32}$

$\therefore$     $\frac{6m+\frac{2}{m}}{\sqrt{m^{2}+1}}=\pm\sqrt{32}$

 $\Rightarrow \left(3m+\frac{1}{m}\right)^{2}=8(m^{2}+1)$

 $\Rightarrow(3m^{2}+1)^{2}=8(m^{4}+m^{2})$

 $\Rightarrow m^{4}-2m^{2}+1=0\Rightarrow m=\pm1$

 Hence, the required tangents are y=x+2 and y=-x-2