If the curves frac{x^{2}}{a}+frac{y^{2}}{b}=1 | vedclass

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(B) Let the point of intersection be $(x_1, y_1)$.For the first curve $\frac{x^2}{a} + \frac{y^2}{b} = 1$,the slope $m_1$ at $(x_1, y_1)$ is given by

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$a-b=c-d$

Vedclass · Answer

(B) Let the point of intersection be $(x_1, y_1)$.For the first curve $\frac{x^2}{a} + \frac{y^2}{b} = 1$,the slope $m_1$ at $(x_1, y_1)$ is given by differentiating: $\frac{2x_1}{a} + \frac{2y_1}{b} \frac{dy}{dx} = 0 \Rightarrow m_1 = -\frac{bx_1}{ay_1}$.For the second curve $\frac{x^2}{c} + \frac{y^2}{d} = 1$,the slope $m_2$ at $(x_1, y_1)$ is $m_2 = -\frac{dx_1}{cy_1}$.Since the curves intersect at $90^{\circ}$,$m_1 m_2 = -1$.$(-\frac{bx_1}{ay_1})(-\frac{dx_1}{cy_1}) = -1 \Rightarrow \frac{bd x_1^2}{ac y_1^2} = -1$,which implies $bd x_1^2 = -ac y_1^2$.Subtracting the two curve equations: $x_1^2(\frac{1}{a} - \frac{1}{c}) + y_1^2(\frac{1}{b} - \frac{1}{d}) = 0$.$x_1^2(\frac{c-a}{ac}) + y_1^2(\frac{d-b}{bd}) = 0$.Substituting $x_1^2 = -\frac{ac}{bd} y_1^2$,we get $(-\frac{ac}{bd}) y_1^2(\frac{c-a}{ac}) + y_1^2(\frac{d-b}{bd}) = 0$.$-\frac{c-a}{bd} + \frac{d-b}{bd} = 0$ $\Rightarrow a-c+d-b = 0$ $\Rightarrow a-b = c-d$.

If the curves $\frac{x^{2}}{a}+\frac{y^{2}}{b}=1$ and $\frac{x^{2}}{c}+\frac{y^{2}}{d}=1$ intersect each other at an angle of $90^{\circ}$,then which of the following relations is $TRUE$?

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