Question

The distance of 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. $a(1+2e\cos \theta )$

Answer 1

The correct option is C $a(1+e\cos \theta )$

Given equation of ellipse is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Any point on the ellipse will be $(acos\theta ,bsin\theta )$

$P=(acos\theta ,bsin\theta )$

Centre of the ellipse is $(0,0)$

Ellipse is parallel to horizontal axis

Foci of the ellipse is

$F=(h-ae,k)$ if $(h,k)$ is the centre

$F=(0-ae,0)=(-ae,o)$

Distance $FP=\sqrt{(-ae-cos\theta {)}^2+(bsin\theta -0{)}^2}$

$=a\sqrt{(e^2+co{s}^2\theta +2ecos\theta +(1-e^2)sin\theta -0{)}^2}$

$a\sqrt{(e^2+co{s}^2\theta +2ecos\theta +si{n}^2\theta -e^{2}si{n}^2\theta )}$

$a\sqrt{(1+2ecos\theta +e^2(1-si{n}^2\theta )}$

$a\sqrt{(1+2ecos\theta +e^{2}co{s}^2\theta )}$

$a\sqrt{(1+ecos\theta {)}^2}$

$FP=a(1+ecos\theta )$