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

$y = 4\sin 3x$ is a solution of the differential equation

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

The differential equation of $y=ae^{bx}$ (where $a$ and $b$ are parameters) is:

The differential equation of a family of hyperbolas whose axes are parallel to coordinate axes, centres lie on the line $y=2x$ and eccentricity is $\sqrt{3}$ is

$A$ differential equation representing the family of parabolas with axis parallel to the $y$-axis and whose length of latus rectum is the distance of the point $(2, -3)$ from the line $3x + 4y = 5$,is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module