Question

The equation of the normal to the curve $y=-\sqrt{x}+2$ at the point of its intersection with the bisector of the first quadrant is :

• A. $2x-y+1=0$
• B. $4x-y+16=0$
• C. $4x-y=16$
• D. $2x-y-1=0$

Answer 1

The correct option is D $2x-y-1=0$


The bisector of the first quadrant is,
$y=x$
Coordinates of $A$ is,
$x=-\sqrt{x}+2\Rightarrow x^2+4-4x=x\Rightarrow x^2-5x+4=0\Rightarrow (x-4)(x-1)=0\Rightarrow x=4\text{or}x=1$

At $x=4$,
$y=-\sqrt{x}+2=0$
This doesn't lie on $y=x$
So $A\equiv (1,1)$

$\frac{dy}{dx}=\frac{-1}{2\sqrt{x}}$
$\frac{dy}{dx}{|}_{(1,1)}=\frac{-1}{2}$

Slope of the normal,
$m_n=-\frac{dx}{dy}$
Slope of the normal at $A$ is, $m_n=2$
Therefore, equation of the normal is,
$y=2x+c$
Putting $(1,1)$
$1=2+c\Rightarrow c=-1$

Hence, the equation of the normal is,
$y=2x-1\Rightarrow 2x-y-1=0$