Q. A curve $y = f(x)$ passes through the point $P(1,\ 1)$. The normal to the curve at point $P$ is $a(y - 1) + (x - 1) = 0$. If the slope of the tangent at any point on the curve is proportional to the ordinate at that point, then the equation of the curve is
A) $y = e^{ax} - 1$
B) $y = e^{ax} + 1$
C) $y = e^{ax} - a$
D) $y = e^{a(x - 1)}$
Explanation
(d)
$\because$ Equation of normal at $P(1,\ 1)$ is
$ay + x = a + 1$ (given)
$\because$ Slope of normal at $(1,\ 1) = - \frac{1}{a}$
$\therefore$ Slope of tangent at $(1,\ 1) = a$ ...(i)
Also, given $\frac{dy}{dx} \propto y$
$\Rightarrow \ \frac{dy}{dx} = ky$
$\left. \ \frac{dy}{dx} \right|_{(1,1)} = k = a$ [from Eq. (i)]
Then, $\frac{dy}{dx} = ay$
$\Rightarrow \frac{dy}{y} = a\ dx$
$\Rightarrow \$In$|y| = ax + c$
$\because$ It is passing through (1,1), then $c = - a$
$\Rightarrow In|y| = a(x - 1)$
$\Rightarrow |y| = e^{a(x - 1)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.