The differential equation of an ellipse whose | vedclass

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

Question

(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Given that the major axis is twice the minor axis,we have $2a = 2(2b)$

Vedclass · Accepted Answer

$x+4 y \frac{dy}{dx}=0$

Vedclass · Answer

(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Given that the major axis is twice the minor axis,we have $2a = 2(2b)$,which implies $a = 2b$. Substituting $a = 2b$ into the equation,we get $\frac{x^2}{(2b)^2} + \frac{y^2}{b^2} = 1$. This simplifies to $\frac{x^2}{4b^2} + \frac{y^2}{b^2} = 1$,or $x^2 + 4y^2 = 4b^2$. Differentiating both sides with respect to $x$,we get $\frac{d}{dx}(x^2 + 4y^2) = \frac{d}{dx}(4b^2)$. This gives $2x + 8y \frac{dy}{dx} = 0$. Dividing by $2$,we obtain $x + 4y \frac{dy}{dx} = 0$.

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

Similar Questions

The differential equation for the line $y = mx + c$ is (where $m$ and $c$ are arbitrary constants).

If the differential equation having $y=Ae^x+B \sin x$ as its general solution is $f(x) \frac{d^2 y}{d x^2}+g(x) \frac{d y}{d x}+h(x) y=0$, then $f(x)+g(x)+h(x)=$

$\tan ^{-1} x + \tan ^{-1} y = c$ is the general solution of the differential equation:

The differential equation of all circles of radius $a$ is of order:

$y = ke^{\sin ^{-1} x} + 3$ is a solution of which differential equation?

Vedclass Test Series

Exam Paper Generator

Online Exam Module