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  • The distance of the point <img alt="\mathrm{\text { ' } \theta^{\prime}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Ctext%20%7B%20%27%20%7D%20%5Ctheta%5E%7B%5Cprime%7D

The distance of the point \mathrm{\text { ' } \theta^{\prime}}  on the ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} from a focus is

Option: 1

\mathrm{a(e+\cos \theta)}


Option: 2

\mathrm{a(e-\cos \theta)}


Option: 3

\mathrm{a(1+e \cos \theta)}


Option: 4

\mathrm{a(1+2 e \cos \theta)}


Answers (1)

best_answer

Focal distance of any point \mathrm{P(x, y)} on the ellipse is equal to \mathrm{S P=a+e x . \text { Here } x=\cos \theta \text {. }}
\mathrm{\text { Hence, } \quad S P=a+a e \cos \theta=a(1+e \cos \theta)}

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Shailly goel

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