Question

The distance of point ${}^{\prime }{\theta }^{\prime }$ 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 as $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Any point on the ellipse will be $(a\cos \theta ,b\sin \theta )$
$\Rightarrow P=(a\cos \theta ,b\sin \theta )$
Center of ellipse $=(0,0)$ and ellipse is parallel to horizontal axis.
$\Rightarrow Fociofellipse,F=(h-ae,k)$( if $(h,k)$ as its center)
$\Rightarrow F=(0-ae,0)=(-ae,0)$
Distance $FP=\sqrt{(-ae-a\cos \theta {)}^2+(b\sin \theta -0{)}^2}$
$=\sqrt{(-ae-a\cos \theta {)}^2+(a\sqrt{(1-e^2)}sin\theta -0{)}^2}$ (since $b=a\sqrt{1-e^2}$)
$=a\times \sqrt{\left(\right(-e-\cos \theta {)}^2+(\sqrt{(1-e^2)}\sin \theta -0{)}^2)}$
$=a\times \sqrt{\left(\right(e^2+{\cos }^2\theta +2e\cos \theta +(1-e^2){\sin }^2\theta \left)\right)}$
$=a\times \sqrt{\left(\right(e^2+{\cos }^2\theta +2e\cos \theta +{\sin }^2\theta -e^2{\sin }^2\theta \left)\right)}$
$=a\times \sqrt{\left(\right(e^2+2e\cos \theta +1-e^2{\sin }^2\theta \left)\right)}$ (since ${\sin }^2\theta +{\cos }^2\theta =1$)
$=a\times \sqrt{\left(\right(1+2e\cos \theta +e^2(1-{\sin }^2\theta )\left)\right)}$
$=a\times \sqrt{\left(\right(1+2e\cos \theta +e^2{\cos }^2\theta \left)\right)}$
$=a\times \sqrt{\left(\right(1+e\cos \theta {)}^2)}$
$FP=a(1+e\cos \theta )$