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

(B) The standard equation of a parabola with focus at $(0,0)$ and axis along the $X$-axis is $(y-0)^2 = 4a(x+a)$,where $a$ is a parameter. This simpli

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The standard equation of a parabola with focus at $(0,0)$ and axis along the $X$-axis is $(y-0)^2 = 4a(x+a)$,where $a$ is a parameter. This simplifies to $y^2 = 4ax + 4a^2$. Differentiating both sides with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies $a = \frac{y}{2} \frac{dy}{dx}$. Substituting this value of $a$ into the original equation: $y^2 = 4\left(\frac{y}{2} \frac{dy}{dx}ight)x + 4\left(\frac{y}{2} \frac{dy}{dx}ight)^2$. $y^2 = 2xy \frac{dy}{dx} + 4\left(\frac{y^2}{4} \left(\frac{dy}{dx}ight)^2ight)$. $y^2 = 2xy \frac{dy}{dx} + y^2 \left(\frac{dy}{dx}ight)^2$. Dividing by $y$ (assuming $y 
eq 0$): $y = 2x \frac{dy}{dx} + y \left(\frac{dy}{dx}ight)^2$. Rearranging the terms gives $-y \left(\frac{dy}{dx}ight)^2 = 2x \frac{dy}{dx} - y$.

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

Similar Questions

Which of the following is a biodegradable polymer?

The locus of the poles of focal chords of the parabola $y^2 = 4ax$ is

Newton's law of cooling is a special case of

Identify the incorrect statement among the following.

Which of the following is not considered an organometallic compound?

Vedclass Test Series

Exam Paper Generator

Online Exam Module