Question

The equation of straight line passing through the point $(3,6)$ and cutting $y=\sqrt{x}$ orthogonally is

• A. $4x+y=18$
• B. $x+y=9$
• C. $4x-y=6$
• D. $x-y=9$

Answer 1

The correct option is A $4x+y=18$

$y=\sqrt{x}\Rightarrow x,y>0$
Now squaring both the sides we get
$y^2=x$
Any line orthogonal to parabolic curve is normal.
So, equation of normal is $y=-tx+a{t}^3+2at$
Here, $a=\frac{1}{4}$
Normal passes through $(3,6)$
$\Rightarrow 6=-3t+\frac{t^3}{4}+\frac{2t}{4}\Rightarrow 24+12t=t^3+2t\Rightarrow t^3-10t-24=0$
By observation, $t=4$ is one solution,
$\Rightarrow (t-4)(t^2+4t+6)=0$

So, $t=4$ is the only real solution
$\therefore$ The required equation of normal is
$y=-4x+\frac{64}{4}+\frac{8}{4}\therefore y+4x=18$