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

Answer 1

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

Given
$y=-\sqrt{x}+\mathrm{2....}\left(i\right)$
And bisector of first quadrant is
$y=x$ ..... $\left(ii\right)$
On solving Eqs(i) and (ii) we get
$x=1,4$
$\therefore$ From Eq (i)
Points are $(1,1)$ and $(4,0)$
But $(4,0)$ not satisfy Eq (ii)
$\therefore$ Point $(1,1)$ is only point of intersection of curve (i) and line (ii)
Now, slope of tangent of curve (i) is
$\frac{dy}{dx}=\frac{-1}{2\sqrt{x}}$
$\therefore$ Slope of tangent at point $(1,1)$ is
${\left|\frac{dy}{dx}\right|}_{(1,1)}=\frac{-1}{2}$
and $\frac{-1}{{\left|\frac{dy}{dx}\right|}_{(1,1)}}=\frac{-1}{\frac{-1}{2}}=2$
$\therefore$ Equation of the normal to the curve at point $(1,1)$ will be
$y-1=2(x-1)\Rightarrow 2x-y-1=0$