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  • For an ellipse, there are exactly how many points of the curve, the normals at which pass through either of the foci.Option: 1 1<div cl

For an ellipse, there are exactly how many points of the curve, the normals at which pass through either of the foci.

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

best_answer

Consider the ellipse \mathrm{\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1}

Equation of normal at any point  \mathrm{(a \cos \theta, b \sin \theta)} is

\mathrm{a \sin \theta \cdot x-b \cos \theta . y=a^2 e^2 \sin \theta \cos \theta}             ....(1)

Take the focus as \mathrm{S(ae, 0).}

Since (1) passes through S(a e, 0), we have

\mathrm{a \sin \theta \cdot a e=a^2 e^2 \sin \theta \cos \theta}

\mathrm{\Rightarrow \sin \theta(1-\mathrm{e} \cos \theta)=0}

\mathrm{\Rightarrow \sin \theta=0} or \mathrm{\cos \theta=\frac{1}{\mathrm{e}}}

\mathrm{\Rightarrow \sin \theta=0 \Rightarrow \theta=0} or \mathrm{\pi \quad(\cos \theta=1 / \mathrm{e}} is not possible )

\Rightarrow corresponding points are \mathrm{(\mathrm{a}, 0)} and \mathrm{(-\mathrm{a}, 0)}

Similarly, if we take \mathrm{S^{\prime} \equiv(-a e, 0)}, then we get same points.

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avinash.dongre

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