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  • Any point   of an ellipse is joined to the extremities of the maj

Any point \mathrm{P}  of an ellipse is joined to the extremities of the major axis. The portion of a  directrix intercepted by them subtends what angle at the corresponding focus ? 

 

Option: 1

75^{\circ}


Option: 2

90^{\circ}


Option: 3

60^{\circ}


Option: 4

120^{\circ}


Answers (1)

best_answer

Let the ellipse be \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 } and \mathrm{\mathrm{P}(\mathrm{a} \cos \phi, b \sin \phi) } be a point on it. The equation of \mathrm{\mathrm{PA} isy-0=\frac{b \sin \phi-0}{a \cos \phi-a}(x-a) }\mathrm{\mathrm{PA} \, \, is\, \, y-0=\frac{b \sin \phi-0}{a \cos \phi-a}(x-a) }

If meets the directrix \mathrm{\mathrm{x}=\frac{\mathrm{a}}{\mathrm{e}} in \mathrm{N} }whose Coordinates are \mathrm{\left[\frac{\mathrm{a}}{\mathrm{e}}, \mathrm{b} \sin \phi \frac{1-\mathrm{e}}{\mathrm{e}(\cos \phi-1)}\right] }
 Equation to \mathrm{ \mathrm{PA}^{\prime} is y=\frac{b \sin \phi}{\mathrm{a} \cos \phi+a}(x+a) }
It meets the directrix \mathrm{ \mathrm{x}=\frac{\mathrm{a}}{\mathrm{e}} \text { in } \mathrm{M} \text { whose\, \, coordinates\, \, are }\left[\frac{\mathrm{a}}{\mathrm{e}}, \mathrm{b} \sin \phi \frac{1+\mathrm{e}}{\mathrm{e}(\cos \phi+1)}\right] }
Now product of gradients of SN and SM
\mathrm{ \begin{aligned} & =\left(\frac{\frac{b \sin \phi(1-e)}{\mathrm{e}(\cos \phi-1)}}{\frac{\mathrm{a}}{\mathrm{e}}-\mathrm{ae}}\right)\left(\frac{\frac{\mathrm{b} \sin \phi(1+\mathrm{e})}{\mathrm{e}(\cos \phi+1)}}{\frac{\mathrm{a}}{\mathrm{e}}-\mathrm{ae}}\right)=\left(\frac{\frac{\mathrm{b} \sin \phi(1-\mathrm{e})}{\mathrm{e}(\cos \phi-1)}}{\frac{\mathrm{a}}{\mathrm{e}}-\mathrm{ae}}\right)\left(\frac{\frac{\mathrm{b} \sin \phi(1+\mathrm{e})}{\mathrm{e}(\cos \phi+1)}}{\mathrm{a}\left(\frac{1}{\mathrm{e}}-\mathrm{e}\right)}\right) \\ & =\left(\frac{\frac{b \sin \phi(1-e)}{\mathrm{e}(\cos \phi-1)} \mathrm{e}}{\mathrm{a}\left(1-\mathrm{e}^2\right)}\right)\left(\frac{\frac{\mathrm{b} \sin \phi(1+\mathrm{e})}{\mathrm{e}(\cos \phi+1)} \mathrm{e}}{\mathrm{a}\left(1-\mathrm{e}^2\right)}\right) \\ & \end{aligned} }

\mathrm{ =\frac{\mathrm{b}^2-\mathrm{e}^2 \sin ^2 \phi}{\left(\cos ^2 \phi-1\right) \mathrm{a}^2\left(1-\mathrm{e}^2\right)^2}=\frac{\mathrm{a}^2\left(1-\mathrm{e}^2\right)\left(1-\mathrm{e}^2\right) \sin ^2 \phi}{-\mathrm{a}^2\left(1-\mathrm{e}^2\right)^2\left(1-\cos ^2 \phi\right)}=-1 }
i.e. \mathrm{ \mathrm{SN} is\, \, perpendicular\, \, to \mathrm{SM} or \mathrm{MN} }subtends a right angle at \mathrm{ \mathrm{S}. }

Posted by

Suraj Bhandari

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