Question

The area in square units bounded by the normal at $(1,2)$ to the parabola $y^2=4x,x$-axis and the curve is given by

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

Answer 1

Answer: (A)

We have, $y^2=4x$
Differentiating w.r.t. ${}^{\prime }{x}^{\prime }$, we get
$2y\frac{dy}{dx}=4\Rightarrow \frac{dy}{dx}=\frac{2}{y}$
${\left(\frac{dy}{dx}\right)}_{\left(1,2\right)}=\frac{2}{2}=1$
Equation of normal to the curve at (1, 2) is
$y-y_1=\frac{1}{\left(-\frac{dy}{dx}\right)}\left(x-x_1\right)$
$\Rightarrow \left(y-2\right)=-\frac{1}{2}\left(x-1\right)$
$\Rightarrow y-2=-x+1$
$\Rightarrow x+y=3$
The line $x+y=3$ meets the $x$-axis at $x=3$
$\therefore$ Required Area = $\underset{0}{\overset{1}{\int }}\sqrt{4x}dx+\underset{1}{\overset{3}{\int }}(3-x)dx$
$=2{\left[\frac{x^{32}}{32}\right]}_0^1+{\left[3x-\frac{x^2}{2}\right]}_1^3$
$=\frac{4}{3}\left(1\right)+\left[9-\frac{9}{2}-3+\frac{1}{2}\right]$
$=\frac{4}{3}+\left(\frac{9}{2}-\frac{5}{2}\right)=\frac{4}{3}+\frac{4}{2}=\frac{4}{3}+2=\frac{10}{3}$ sq. units