If the curves frac{x^2}{a^2} + frac{y^2}{b^2} | vedclass

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(D) For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,differentiating with respect to $x$ gives $\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} =

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$a^2 + b^2 = l^2 - m^2$

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(D) For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,differentiating with respect to $x$ gives $\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{xb^2}{ya^2}$.For the hyperbola $\frac{x^2}{l^2} - \frac{y^2}{m^2} = 1$,differentiating with respect to $x$ gives $\frac{2x}{l^2} - \frac{2y}{m^2} \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = \frac{xm^2}{yl^2}$.Since the curves cut orthogonally,the product of their slopes is $-1$:$(-\frac{xb^2}{ya^2}) 	imes (\frac{xm^2}{yl^2}) = -1 \Rightarrow \frac{x^2}{y^2} = \frac{a^2 l^2}{b^2 m^2}$.Subtracting the two curve equations: $\frac{x^2}{a^2} + \frac{y^2}{b^2} - (\frac{x^2}{l^2} - \frac{y^2}{m^2}) = 1 - 1 = 0$.$x^2(\frac{1}{a^2} - \frac{1}{l^2}) + y^2(\frac{1}{b^2} + \frac{1}{m^2}) = 0 \Rightarrow \frac{x^2}{y^2} = -\frac{(\frac{1}{b^2} + \frac{1}{m^2})}{(\frac{1}{a^2} - \frac{1}{l^2})} = \frac{(m^2 + b^2)a^2 l^2}{b^2 m^2 (l^2 - a^2)}$.Equating the two expressions for $\frac{x^2}{y^2}$:$\frac{a^2 l^2}{b^2 m^2} = \frac{(m^2 + b^2)a^2 l^2}{b^2 m^2 (l^2 - a^2)} \Rightarrow 1 = \frac{m^2 + b^2}{l^2 - a^2}$.$l^2 - a^2 = m^2 + b^2 \Rightarrow a^2 - b^2 = l^2 - m^2$ is incorrect; re-evaluating: $l^2 - a^2 = m^2 + b^2 \Rightarrow a^2 - b^2 = l^2 - m^2$ is not the standard form. Let's re-check: $l^2 - a^2 = m^2 + b^2 \Rightarrow a^2 + b^2 = l^2 - m^2$.

If the curves $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and $\frac{x^2}{l^2} - \frac{y^2}{m^2} = 1$ cut each other orthogonally,then :-

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