Question

The parametric representation of a point on the ellipse whose foci are $(-1,0)$ and $(7,0)$ and eccentricity $\frac{1}{2}$ is:

• A. $(3+8\cos \theta ,4\sqrt{3}\sin \theta )$
• B. $(8\cos \theta ,4\sqrt{3}\sin \theta )$
• C. $(3+4\sqrt{3}\cos \theta ,8\sin \theta )$
• D. None of these

Answer 1

Answer: (A)

$(3+8\cos \theta ,4\sqrt{3}\sin \theta )$

Using Midpoint formula $X=\left(\frac{x_1+x_2}{2}\right)andY=\left(\frac{y_1+y_2}{2}\right)$

Center of the ellipse is the mid point of foci i.e $(\frac{-1+7}{2}+\frac{0+0}{2})$ which is $(3,0)$

Distance Formula $=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Now, Distance between the foci $=\sqrt{(7+1{)}^2+0}=8=2ae\Rightarrow ae=4\Rightarrow a=8$ since $e=\frac{1}{2}$ (given)

So $b=8\sqrt{1-e^2}=8\sqrt{1-1/4}=4\sqrt{3}$

Hence equation of ellipse is given by,

$\Rightarrow \frac{(x-3{)}^2}{8^2}+\frac{(y-0{)}^2}{(4\sqrt{3}{)}^2}=1$

So parametric points is, $x-3=8\cos \theta$ and $y-0=4\sqrt{3}\sin \theta$

$\Rightarrow x=3+8\cos \theta ,y=4\sqrt{3}\sin \theta$