Question

Gorm the differential equation of an ellipse having foci at x axis and

Answer 1

The line segment of length 2b perpendicular to the transverse axis whose midpoint is the center is the conjugate axis of the hyperbola. The standard equation for a hyperbola with a vertical transverse axis is - = 1. The center is at (h, k). The distance between the vertices is 2a.

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Answer 2

ANSWER:-

We know that the equation of said family of ellipses is :-

$\sf\red{\frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = 1.........(i)}$

Differentiating equation (i) with respect to x, we get $\sf\red{\frac{2x}{ {a}^{2} } + \frac{2y}{ {b}^{2} } \frac{dy}{dx} = 0}$

Or

$\sf\red{\frac{y}{x} ( \frac{dy}{dx} ) = \frac{ { - b}^{2} }{ {a}^{2} } ........(ii) }$

Differentiating both sides of equation (ii) with respect to x,we get

$\longrightarrow$$\sf\red{( \frac{y}{x} )( \frac{ {d}^{2}y }{ {dx}^{2} } ) + ( \frac{x \frac{dy}{dx} - y}{ {x}^{2} }) \frac{dy}{dx} = 0}$

$\longrightarrow$$\sf\red{xy \frac{ {d}^{2} y}{ {dx}^{2} } + x {( \frac{dy}{dx}) }^{2} - y \frac{dy}{dx} = 0.......(iii)}$

which is the required differential equation.