The differential equation of all parabolas wh | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The equation of a parabola with its axis parallel to the $y$-axis is given by $(x-h)^2 = 4a(y-k)$. Since the axis is the $y$-axis,the vertex lies

Vedclass · Accepted Answer

$x \frac{d^2 y}{d x^2}-\frac{d y}{d x}=0$

Vedclass · Answer

(C) The equation of a parabola with its axis parallel to the $y$-axis is given by $(x-h)^2 = 4a(y-k)$. Since the axis is the $y$-axis,the vertex lies on the $y$-axis,so $h=0$. Thus,the equation is $x^2 = 4a(y-k)$.Differentiating with respect to $x$:$2x = 4a \frac{dy}{dx} \implies 4a = \frac{2x}{dy/dx}$.Differentiating again with respect to $x$:$2 = 4a \frac{d^2y}{dx^2}$.Substituting the value of $4a$ from the first derivative into the second derivative equation:$2 = \left( \frac{2x}{dy/dx} \right) \frac{d^2y}{dx^2}$.Simplifying:$1 = \frac{x}{dy/dx} \cdot \frac{d^2y}{dx^2}$$\implies \frac{dy}{dx} = x \frac{d^2y}{dx^2}$$\implies x \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0$.

The differential equation of all parabolas whose axis is the $y$-axis is:

Similar Questions

The differential equation satisfied by the family of curves $y = ax \cos \left( \frac{1}{x} + b \right)$,where $a$ and $b$ are parameters,is

Find the differential equation for the family of curves given by eliminating arbitrary constants $a$ and $b$ from the equation: $\frac{x}{a} + \frac{y}{b} = 1$.

The differential equation satisfied by the function $y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \dots \infty}}}$ is

The differential equation whose solution is $y=c^2+\frac{c}{x}$,where $c$ is a constant,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module