Question

Find the equations of tangents to parabola $y^2=4ax$ which are drawn from the point (2a,3a).

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

Answer 1

The correct option is A $x-y+a=0,x-2y+4a=0$

Any tangent to parabola is given by, $y=mx+\frac{a}{m}$
Now given it passes through $(2a,3a)$
$\Rightarrow 3a=2am+\frac{a}{m}\Rightarrow 2m^2-3m+1=0\Rightarrow m=1,\frac{1}{2}$
Hence equation of tangents are, $(y-3a)=1(x-2a)$ and $(y-3a)=\frac{1}{2}(x-2a)$
$\Rightarrow x-y+a=0$ and $x-2y+4a=0$