Question

A curve $y=f\left(x\right)$ passes through the point $P(1,1).$ The normal to the curve at $P$ is $(y-1)+(x-1)=0.$ If the slope of the tangent at any point on the curve is proportional to the ordinate of the point, then the equation of the curve is:

• A. $y=e^{2(x-1)}$
• B. $ey=e^{3-2x}$
• C. $y=e^{2(1-x)}$
• D. $y=e^{(x-1)}$

Answer 1

The correct option is D $y=e^{(x-1)}$

Slope of the normal at $(1,1)=-1$
So, the slope of the tangent at $(1,1)=1$
i.e., ${\left(\frac{dy}{dx}\right)}_{(1,1)}=1$
Since $\frac{dy}{dx}$ is proportional to $y,$
$\therefore \frac{dy}{dx}=Ky$
Here $K=1\because {\left(\frac{dy}{dx}\right)}_{(1,1)}=1$
$\Rightarrow \frac{dy}{y}=dx$
$\Rightarrow {\log }_{e}y=x+C$
$\Rightarrow y=e^{x+C}=A{e}^x$ where $A=e^c$
It passes through $(1,1)$
$\therefore 1=Ae\Rightarrow A=e^{-1}$
$\therefore y=e^{-1}{e}^x=e^{(x-1)}$