Question

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

• A. $4x+y-18=0$
• B. $x+y-9=0$
• C. $4x-y-6=0$
• D. none of these

Answer 1

Answer: (A)

$4x+y-18=0$

The curve $y=\sqrt{x}$ is the part of curve $y^2=x$
Equation of normal at $P\left(\frac{t^2}{4},\frac{t}{2}\right)$ is
$y+tx=\frac{t}{2}+\frac{t^3}{4}$ ..... (1)
The equation will pass through $(3,6)$
$6+3t=\frac{t}{2}+\frac{t^3}{4}$
$t^3-10t-24=0$
Solving, we get $t=4$
So equation of line which is orthogonal and passes through $(3,6)$ is
$y+4x=18$