Differential Equations
Differential Equation
nta_abhyas_2025
Grade 12
Question:
Tangent to a curve intersects the $y$-axis at point $P$. A line perpendicular to this tangent through $P$ passes through the point $(1, 0)$. The differential equation of the curve is
y' = -x\left(\frac{dy}{dx}\right)^2 = 1
x\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 1
y\frac{d^2y}{dx^2} + x = 1
None of these
Step-by-Step Solution
Key Concept: The perpendicular from a point to a tangent line involves using the negative reciprocal of the slope.
The equation of tangent at point $(x, f(x))$ is $R: (s, f(s)) = y - f(x) = f'(x)(X - x)$. The coordinates of point $P$ where the perpendicular from $P$ to the tangent meets it is given. The slope of the perpendicular line through $P$ is $\frac{f(x) - f(s)}{-1} = \frac{1}{f'(x)}$, which leads to $y\frac{dy}{dx} - x\left(\frac{dy}{dx}\right)^2 = 1$ as the differential equation.
Correct Answer: 1