Question

The area bounded in the first quadrant by the normal at $(1,2)$ on the curve $y^2=4x$, x-axis and the curve is given by$:$

• A. $\frac{10}{3}$
• B. $\frac{7}{3}$
• C. $\frac{4}{3}$
• D. $\frac{9}{2}$

Answer 1

Answer: (A)

$\frac{10}{3}$

Consider the given curve.

$y^2=4x$

Slope of tangent at $(1,2)$ is,

$\frac{dy}{dx}=\frac{2}{y}=1$

Slope of normal at $(1,2)$ is,

$\Rightarrow \frac{-1}{1}=-1$

Therefore,
Equation of normal is: $x+y-3=0$

Normal intersects x-axis at $(3,0)$

Therefore,
$\begin{array}{c}Area={\int }_0^{1}2\sqrt{x}+\frac{1}{2}\times 2\times 2\\ =2\times \frac{{\left[x^{3/2}\right]}_0^1}{3/2}+1\\ =\frac{4}{3}+2=\frac{10}{3}\end{array}$