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 equal to

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

Answer 1

The correct option is A $\frac{10}{3}$ sq. units

We have,
$y^2=4x$
$\Rightarrow \frac{dy}{dx}=\frac{4}{2y}\Rightarrow {\frac{dy}{dx}|}_{(1,2)}=1$
$\therefore$ Equation of normal:
$\frac{y-2}{x-1}=-1$
$\Rightarrow y-2=1-x$
$\Rightarrow x+y=3\cdots \left(i\right)$
Now,
Required area
$=\underset{0}{\overset{1}{\int }}\sqrt{4x}dx+\frac{1}{2}\times 2\times 2=2\times \frac{2}{3}{\left[x^{\frac{3}{2}}\right]}_0^1+2=\frac{4}{3}+2$
$=\frac{10}{3}$ sq. units