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प्रश्न
Show that the curves \[\frac{x^2}{a^2 + \lambda_1} + \frac{y^2}{b^2 + \lambda_1} = 1 \text { and } \frac{x^2}{a^2 + \lambda_2} + \frac{y^2}{b^2 + \lambda_2} = 1\] intersect at right angles ?
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उत्तर
\[\text { We have, } \frac{x^2}{a^2 + \lambda_1} + \frac{y^2}{b^2 + \lambda_1} = 1 . . . \left( 1 \right)\]
\[\text { and } \frac{x^2}{a^2 + \lambda_2} + \frac{y^2}{b^2 + \lambda_2} = 1 . . . \left( 2 \right)\]
\[\text { Now we can find the slope of both the curve by differentiating w . r . t x }\]
\[ \Rightarrow \frac{2x}{a^2 + \lambda_1} + \frac{2y\frac{dy}{dx}}{b^2 + \lambda_1} = 0 \text { and } \frac{2x}{a^2 + \lambda_2} + \frac{2y\frac{dy}{dx}}{b^2 + \lambda_2} = 0\]
\[ \Rightarrow \frac{dy}{dx} = - \frac{x}{y} \times \frac{b^2 + \lambda_1}{a^2 + \lambda_1} \text { and } \frac{dy}{dx} = - \frac{x}{y} \times \frac{b^2 + \lambda_2}{a^2 + \lambda_2}\]
\[ \Rightarrow m_1 = - \frac{x}{y} \times \frac{b^2 + \lambda_1}{a^2 + \lambda_1} \text { and } m_2 = - \frac{x}{y} \times \frac{b^2 + \lambda_2}{a^2 + \lambda_2}\]
\[\text { Subtracting} \left( 2 \right) \text { from } \left( 1 \right), \text { we get }, \]
\[ x^2 \left( \frac{1}{a^2 + \lambda_1} - \frac{1}{a^2 + \lambda_2} \right) + y^2 \left( \frac{1}{b^2 + \lambda_1} - \frac{1}{b^2 + \lambda_2} \right) = 0\]
\[ \Rightarrow \frac{x^2}{y^2} = \frac{\lambda_2 - \lambda_1}{\left( b^2 + \lambda_1 \right)\left( b^2 + \lambda_2 \right)} \times \frac{1}{\frac{\lambda_1 - \lambda_2}{\left( a^2 + \lambda_1 \right)\left( a^2 + \lambda_2 \right)}}\]
\[\text { Now,} \]
\[ m_1 \times \times m_2 = \frac{x^2}{y^2} \times \frac{b^2 + \lambda_1}{a^2 + \lambda_1} \times \frac{b^2 + \lambda_2}{a^2 + \lambda_2}\]
\[ = \frac{\lambda_2 - \lambda_1}{\left( b^2 + \lambda_1 \right)\left( b^2 + \lambda_2 \right)} \times \frac{\left( a^2 + \lambda_1 \right)\left( a^2 + \lambda_2 \right)}{\lambda_1 - \lambda_2} \times \frac{b^2 + \lambda_1}{a^2 + \lambda_1} \times \frac{b^2 + \lambda_2}{a^2 + \lambda_2}\]
\[ = - 1\]
\[\text { hence,} \left( 1 \right) \text { and } \left( 2 \right) \text { cuts orthogonally } . \]
