A curve $f(x)$ 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
Step-by-Step Solution
Key Concept: The normal's slope determines the tangent's slope via perpendicularity: if normal has slope $-\frac{1}{a}$, then tangent has slope $a$. This equals the proportionality constant in the differential equation $\frac{dy}{dx} = ky$, giving $k = a$. Solving the separable ODE $\frac{dy}{dx} = ay$ with initial condition $y(1) = 1$ yields $\ln y = a(x-1)$ or $y = e^{a(x-1)}$.
The normal at point $P(1,1)$ has equation $ay + x = a + 1$, giving slope of tangent $a$. Given $\frac{dy}{dx} = ky$, we separate variables to get $\ln y = ax + c$. Using the initial condition that the curve passes through $(1,1)$, we find $c = -a$. Therefore, the equation of the curve is $y = e^{a(x-1)}$.
Correct Answer: 3