The differential equation of the family of pa | vedclass

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

Question

(A) The equation of a parabola with focus at the origin $(0,0)$ and the $X$-axis as its axis is given by $(y-0)^2 = 4a(x+a)$,which simplifies to $y^2

Vedclass · Accepted Answer

$-y\left(\frac{dy}{dx}\right)^2 = 2x\frac{dy}{dx} - y$

Vedclass · Answer

(A) The equation of a parabola with focus at the origin $(0,0)$ and the $X$-axis as its axis is given by $(y-0)^2 = 4a(x+a)$,which simplifies to $y^2 = 4a(x+a)$.Differentiating both sides with respect to $x$,we get $2y\frac{dy}{dx} = 4a$,which implies $a = \frac{y}{2}\frac{dy}{dx}$.Substituting the value of $a$ into the original equation $y^2 = 4a(x+a)$,we get:$y^2 = 4\left(\frac{y}{2}\frac{dy}{dx}ight)\left(x + \frac{y}{2}\frac{dy}{dx}ight)$$y^2 = 2y\frac{dy}{dx}\left(x + \frac{y}{2}\frac{dy}{dx}ight)$Dividing by $y$ (assuming $y 
eq 0$):$y = 2x\frac{dy}{dx} + y\left(\frac{dy}{dx}ight)^2$Rearranging the terms,we get $y\left(\frac{dy}{dx}ight)^2 + 2x\frac{dy}{dx} - y = 0$,which is equivalent to $y\left(\frac{dy}{dx}ight)^2 = y - 2x\frac{dy}{dx}$ or $-y\left(\frac{dy}{dx}ight)^2 = 2x\frac{dy}{dx} - y$.

The differential equation of the family of parabolas with focus at the origin and the $X$-axis as axis,is

Similar Questions

The differential equation of the family of circles passing through the points $(0,2)$ and $(0,-2)$ is

The differential equation representing the family of circles having their centres on the $Y$-axis is (where $y_1 = \frac{dy}{dx}$ and $y_2 = \frac{d^2y}{dx^2}$):

The differential equation of the family of curves $r^2 = a^2 \cos 2\theta$,where '$a$' is an arbitrary constant,is:

The differential equation of $y = \sec(\tan^{-1}x)$ is

The differential equation of an ellipse whose major axis is twice its minor axis,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module