The differential equation of a family of hype | vedclass

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(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

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

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(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

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