Question

find dy/dx,. if x^2/a^2 + y^2/b^2 = xy​

Answer 1

Answer:

$\frac{dy}{dx} = \frac{a^2b^2y - 2b^2x}{2a^2y - a^2b^2x}$

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Step-by-step explanation:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = xy$

$or, b^2x^2+a^2y^2 = a^2b^2xy$

Differentiating both side w.r.t.x,

$or, \frac{d}{dx} (b^2x^2 +a^2y^2) = \frac{d}{dx}a^2b^2xy\\or, 2b^2x+a^2\frac{dy^2}{dy} * \frac{dy}{dx} = a^2b^2[y \frac{dx}{dx} + x \frac{dy}{dx}]\\ or,2b^2x + 2a^2y* \frac{dy}{dx} = a^2b^2y + a^2b^2x \frac{dy}{dx}\\ or,2a^2y \frac{dy}{dx}-a^2b^2x \frac{dy}{dx} = a^2b^2y - 2b^2x\\or, [2a^2y -a^2b^2x]\frac{dy}{dx} =a^2b^2y - 2b^2x\\or,\frac{dy}{dx } =\frac{a^2b^2y - 2b^2x}{2a^2y -a^2b^2x}$