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If \mathrm{\theta \text { and } \phi} and are eccentric angles of the ends of a pair of conjugate diameters of the ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} ,then \mathrm{\theta-\phi} is equal to

Option: 1

\mathrm{\pm \frac{\pi}{2}}


Option: 2

\mathrm{\pm \pi}


Option: 3

0


Option: 4

None of these


Answers (1)

best_answer

Let , \mathrm{y=m_1 x \text { and } y=m_2 x} be a pair of conjugate diameter of an ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} and let \mathrm{P(a \cos \theta, b \sin \theta)} and \mathrm{Q(a \cos \phi, b \sin \phi)} be ends of these two diameters. Then \mathrm{m_1 m_2=\frac{-b^2}{a^2}} 
\mathrm{\left.\Rightarrow \quad \frac{b \sin \theta-0}{a \cos \theta-0} \times \frac{b \sin \phi-0}{a \cos \phi-0}=\frac{-b^2}{a^2} \Rightarrow \sin \theta \sin \phi=-\cos \theta \cos \phi \Rightarrow \cos \theta-\phi\right)=0 \Rightarrow \theta-\phi= \pm \pi / 2}

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