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 the family of curves $y=e^{x}(A \cos x+B \sin x)$,where $A$ and $B$ are arbitrary constants,is

The general solution of the differential equation of all circles having center at $A(-1, 2)$ is $ . . . . . . $.

The degree of the differential equation whose solution is $y^2=8a(x+a)$ is

The differential equation of all parabolas,whose axes are parallel to the $Y$-axis,is

The differential equation whose solution is $Ax^2 + By^2 = 1$,where $A$ and $B$ are arbitrary constants,is of

Vedclass Test Series

Exam Paper Generator

Online Exam Module