Distance of point 'θ ' on ellipse frac{{{x^2}}}{{{a^2}}} + frac{{{y^2}}}{{{b^2}}} = 1 from

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10-2. Parabola, Ellipse, Hyperbola

easy

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

A

$a(e + \cos \theta )$

B

$a(e - \cos \theta )$

C

$a(1 + e\cos \theta )$

D

$a(1 + 2e\cos \theta )$

Solution

(c) Focal distance of any point $P (x,y)$ on the ellipse is equal to $SP = a + ex$.

Here $x = a\cos \theta $

Here $SP = a + ae\cos \theta = a(1 + e\cos \theta )$.

Standard 11

Mathematics

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