Differential Equations
Differential Equations
star_batch_jee_advanced_2025
Grade 12
Question:
If a tangent drawn to the curve $y = f(x)$ at $(x, y)$ cuts the $x-$ axis and $y-$ axis at $A$ and $B$ respectively such that $\frac{BP}{AP} = \frac{3}{1}$ given $f(1) = 1$, then:
differential equation of curve may be $x\frac{dy}{dx}+3y = 0$
differential equation of curve may be $x\frac{dy}{dx} = 3y$
if tangent at $R(a, b)$ intersect again at $S(m,n)$ then $m+2a = 0$
equation of normal at $(1,1)$ is $3y = x+2$
Step-by-Step Solution
Key Concept: Geometric conditions on tangent and normal lines translate into differential equations through the relationship between slopes and coordinates.
For a curve with tangent and normal equations given, point $A$ is the x-intercept of the tangent at $P(x,y)$: $A = (x - y\frac{dx}{dy}, 0)$. Point $B$ is the y-intercept of the normal: $B = (0, y - x\frac{dy}{dx})$. The slopes give $\frac{BP}{AP} = \frac{|y|}{|x|}\cdot\frac{|dy/dx|}{|dy/dx|} = \frac{3}{1}$, leading to $\frac{dy}{dx} + \frac{3y}{x} = 0$. Solving this linear equation: $\ln y = 3\ln x^{-1}$, so $y = \frac{1}{x^3}$.
Correct Answer: 1,2,3,4