Question

Parametric coordinates of a point on ellipse, whose foci are $(-1,0)$ and $(7,0)$ and eccentricity is $\frac{1}{2},$ is

• A. $(8+3\cos \theta ,4\sqrt{3}\sin \theta )$
• B. $(8-3\cos \theta ,4\sqrt{3}\sin \theta )$
• C. $(3+8\cos \theta ,-2\sqrt{3}\sin \theta )$
• D. $(3+8\cos \theta ,4\sqrt{3}\sin \theta )$

Answer 1

The correct option is D $(3+8\cos \theta ,4\sqrt{3}\sin \theta )$

Let the equation of ellipse is $\frac{(x-x_0{)}^2}{a^2}+\frac{(y-y_0{)}^2}{b^2}=1$
where $(x_0,y_0)$ is the centre of ellipse
Foci are $(-1,0)$ and $(7,0)$.
Distance between foci is $2ae=7-(-1)=8$.
$\Rightarrow ae=4$
since $e=\frac{1}{2}\Rightarrow a=8.$
Now, $b^2=a^2(1-e^2)\Rightarrow b=\sqrt{8^2(1-{\left(\frac{1}{2}\right)}^2)}=4\sqrt{3}.$

The centre of the ellipse is the midpoint of the line joining two foci,
$\therefore$ Centre $\equiv (\frac{-1+7}{2},\frac{0+0}{2})$
$\Rightarrow$ Centre $\equiv (3,0)$
So, equation of the ellipse is $\frac{(x-3{)}^2}{8^2}+\frac{(y-0{)}^2}{(4\sqrt{3}{)}^2}=1$
Hence, the parametric coordinates of a point is $(x_0+a\cos \theta ,y_0+b\sin \theta )=(3+8\cos \theta ,4\sqrt{3}\sin \theta ).$