Question

The area bounded by the curve $y=f\left(x\right)$, y-axis and the normal to the curve at $P$ is

• A. $\frac{1}{a}\left(\begin{array}{c}a-\frac{1}{2}+e^{-a}\end{array}\right)$
• B. $\frac{1}{a}\left(\begin{array}{c}a+\frac{1}{2}+e^{-a}\end{array}\right)$
• C. $\frac{1}{a}\left(\begin{array}{c}a-\frac{1}{2}+e^a\end{array}\right)$
• D. $\frac{1}{a}\left(\begin{array}{c}a+\frac{1}{2}+e^a\end{array}\right)$

Answer 1

The correct option is A $\frac{1}{a}\left(\begin{array}{c}a-\frac{1}{2}+e^{-a}\end{array}\right)$

$Fig.$$\left(1\right)$

Given:

curve $y=f\left(x\right)$ passes through the point $P(1,1)$ ;

Normal at the point $P(1,1)$ is $a(y-1)+(x-1)=0$ ;

slope of tangent $\frac{dy}{dx}\propto y$ (ordinate of point )

$\Rightarrow$ $\frac{dy}{dx}=ky$ where $k$ is a proportionality constant.

$\Rightarrow$ $\frac{dy}{y}=k.dx$

integrating both the sides

$\int \frac{dy}{y}=\int k.dx$

$\Rightarrow$ $ln\left(y\right)=kx+ln\left(c\right)$ where ${}^{\prime }ln(c{)}^{\prime }$ is a integral constant

$\Rightarrow$ $ln\left(\frac{y}{c}\right)=kx$

$\Rightarrow$ $\frac{y}{c}=e^{kx}$

$\Rightarrow$ $y=c.e^{kx}$

curve $y=c.e^{kx}$ passes through point $P(1,1)$

$\Rightarrow$ $1=c.e^{k.1}$

$\Rightarrow$ $1=c.e^k$

$\Rightarrow$ $c=e^{-k}$

on putting this value of ${}^{\prime }{c}^{\prime }$ into the equation of curve

$y=e^{-k}.e^{kx}$

$\Rightarrow$ $y=e^{k(x-1)}$

Slope of $Normal$ at point $P(1,1)$

$N_{(1,1)}$ $=$ $-\frac{dx}{dy}$

$\Rightarrow$ $N_{(1,1)}$ $=$ $-\frac{1}{\left(\frac{dy}{dx}\right)}$

$\Rightarrow$ $N_{(1,1)}$ $=$$-\frac{1}{\frac{d}{dx}\left(e^{k(x-1)}\right)}$

$\Rightarrow$ $N_{(1,1)}$ $=$$-\frac{1}{k.e^{(1-1)}}$

$\Rightarrow$ $N_{(1,1)}$ $=$$-\frac{1}{k.e^{\left(0\right)}}$

$\Rightarrow$ $N_{(1,1)}$ $=$ $-\frac{1}{k}$ $......\left(1\right)$

equation of $Normal$ at point $P(1,1)$ is $a(y-1)+(x-1)=0$

slope of normal will be $\frac{dy}{dx}$ of the equation of normal at point $P(1,1)$

$a(y-1)+(x-1)=0$

$\Rightarrow$ $a(\frac{dy}{dx}-0)+(1-0)=0$

$\Rightarrow$ $\frac{dy}{dx}=-\frac{1}{a}$

So $N_{(1,1)}$ $=$ $-\frac{1}{a}$ $......\left(2\right)$

from equation $\left(1\right)$ and $\left(2\right)$

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

$\Rightarrow$ $a=k$

Now equation of curve $y=e^{a(x-1)}$

equation of normal $a(y-1)+(x-1)=0$

$y=e^{a(x-1)}$

$\Rightarrow$ at $x=0$, $y=e^{-a}$

$\Rightarrow$ at $x=1$, $y=1$

$a(y-1)+(x-1)=0$

$\Rightarrow$ at $x=0$, $y=(1+\frac{1}{a})$

$\Rightarrow$ at $x=1$, $y=1$

Area bounded by the curve $y=f\left(x\right)$ , $y-axis$ and the normal to the curve at point $P(1,1)$

Total area $=$ shaded $area\left(1\right)$ + shaded $area\left(2\right)$; $(shown$ $in$ $Fig.$ $1)$

Now, $shaded$ $area\left(1\right)$ $=$ area of triangle having coordinates $(0,1)$, $(1,1)$, $(0,1+$ $\frac{1}{a}$)

$=$ $\frac{1}{2}\times \left(base\right)\times \left(height\right)$

$=$ $\frac{1}{2}\times (1-0)\times (1$+$\frac{1}{a}-1$)

$=$ $\frac{1}{2}\times \left(1\right)\times$$(\frac{1}{a}$)

$=$ $\frac{1}{2a}$

so $shaded$ $area\left(1\right)$ $=$$\frac{1}{2a}$

$\because$ $y=e^{a(x-1)}$

$\Rightarrow$ $ln\left(y\right)=a(x-1)$

$\Rightarrow$ $x=$ $\frac{1}{a}$$ln\left(y\right)+1$

Now,

$shaded$ $area\left(2\right)$ $=$ ${\int }_{e^{-a}}^{1}xdy$

$=$ ${\int }_{e^{-a}}^1$ ( $\frac{1}{a}$$ln\left(y\right)+1)$ $dy$

$=$$\frac{1}{a}{\int }_{e^{-a}}^{1}ln\left(y\right)dy$ $+$ ${\int }_{e^{-a}}^{1}1dy$

$=$ $\frac{1}{a}\times [y.ln(y)-y{]}_{e^{-a}}^1$ $+$ $[y{]}_{e^{-a}}^1$

$=$ $\frac{1}{a}[-1+a{e}^{-a}+e^{-a}]$ $+$ $[1-e^{-a}]$

$=$ $(1-\frac{1}{a}+\frac{e^{-a}}{a})$

so $shaded$ $area\left(2\right)$ $=$ $(1-\frac{1}{a}+\frac{e^{-a}}{a})$

$Total$ $area$ $=$ $shaded$ $area\left(1\right)$ + $shaded$ $area\left(2\right)$

$=$ $\left(\frac{1}{2a}\right)$ $+$ $(1-\frac{1}{a}+\frac{e^{-a}}{a})$

$=$ $(1-\frac{1}{2a}+\frac{e^{-a}}{a})$

$=$ $\frac{1}{a}(a-\frac{1}{2}+e^{-a})$

Hence the correct answer is $\frac{1}{a}(a-\frac{1}{2}+e^{-a})$ is correct.