The differential equation of a family of hype | vedclass

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

Question

(B) The equation of a hyperbola with axes parallel to the coordinate axes and center $(h, k)$ is given by $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} =

Vedclass · Accepted Answer

$(y-2x)y_2+y_1^2+2y_1=y_1^3+2$

Vedclass · Answer

(B) The equation of a hyperbola with axes parallel to the coordinate axes and center $(h, k)$ is given by $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ or $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$. Given the center lies on $y=2x$, we have $k=2h$. For a hyperbola, $e^2 = 1 + \frac{b^2}{a^2} = 3$, so $b^2 = 2a^2$. The equation becomes $(x-h)^2 - \frac{1}{2}(y-2h)^2 = \pm a^2$. Differentiating with respect to $x$: $2(x-h) - (y-2h)y_1 = 0$, which gives $x-h = \frac{1}{2}(y-2h)y_1$. Substituting $h = x - \frac{1}{2}(y-2h)y_1$ into the equation and simplifying leads to the differential equation $(y-2x)y_2 + y_1^2 + 2y_1 = y_1^3 + 2$.

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

Similar Questions

The differential equation of all circles which pass through the origin and whose centres lie on the $y$-axis is:

The differential equation for which $\sqrt{1+y^2}=C x e^{\tan ^{-1} x}$ is the general solution, is

Verify that the function $y=a \cos x+b \sin x$,where $a, b \in \mathbb{R}$,is a solution of the differential equation $\frac{d^{2} y}{d x^{2}}+y=0$.

For each of the exercises given below,verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
$y=e^{x}(a \cos x+b \sin x) \quad: \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0$

Verify that the given function $y=e^{x}+1$ is a solution of the differential equation $y^{\prime \prime}-y^{\prime}=0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module