Question

The parametric representation of a point of the ellipse whose foci are $\left(-3,0\right)$ and $\left(9,0\right)$ and eccentricity $\frac{1}{3}$ is

• A. $\left(3+18\cos \theta ,12\sqrt{2}\sin \theta \right)$
• B. $\left(1+3\cos \theta ,5\sin \theta \right)$
• C. $\left(1+3\cos \theta ,1+\sqrt{5}\sin \theta \right)$
• D. $\left(1+3\cos \theta ,1+5\sin \theta \right)$
• E. $\left(1+3\cos \theta ,\sqrt{5}\sin \theta \right)$

Answer 1

The correct option is A

$\left(3+18\cos \theta ,12\sqrt{2}\sin \theta \right)$

Step 1: Calculate the distance between the foci

The foci of the ellipse are $\left(-3,0\right)$ and $\left(9,0\right)$.

And, the eccentricity $\left(e\right)=\frac{1}{3}$.

Now, using the distance formula, the distance between the foci can be calculated as,

The distance between the foci $=\sqrt{{\left(9+3\right)}^2+{\left(0+0\right)}^2}$

$\Rightarrow$ The distance between the foci $=\sqrt{{\left(12\right)}^2+{\left(0\right)}^2}$

$\Rightarrow$The distance between the foci $=12\mathrm{units}$ $...\left(\mathrm{i}\right)$

Step 2: Calculate the value of $a$

As we know,

The distance between the foci of an ellipse $=2ae$

$\Rightarrow$ $12=2ae$ [using equation $\left(\mathrm{i}\right)$]

$\Rightarrow$ $12=2a\times \frac{1}{3}$

$\Rightarrow$ $12\times \frac{3}{2}=a$

$\Rightarrow$ $18=a$

Step 3: Calculate the value of $b$

Again, we know that,

$e=\sqrt{1-\frac{b^2}{a^2}}$

$\Rightarrow$ $\frac{1}{3}=\sqrt{1-\frac{b^2}{{\left(18\right)}^2}}$ $\left[\because a=18,e=\frac{1}{3}\right]$

$\Rightarrow$ $\frac{1}{9}=1-\frac{b^2}{324}$ [squaring on both sides]

$\Rightarrow$ $\frac{b^2}{324}=1-\frac{1}{9}$

$\Rightarrow$ $b^2=\frac{8}{9}\times 324$

$\Rightarrow$ $b^2=288$

$\Rightarrow$ $b=12\sqrt{2}$

Step 4: Calculate the coordinates of the center of the ellipse

As we know, the center of the ellipse is the mid point of the line obtained by joining the foci of the ellipse.

Also, by the mid-point formula, the mid-point of a line obtained by joining two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by,

$\left(x,y\right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

As given, the foci of the ellipse are $\left(-3,0\right)$ and $\left(9,0\right)$.

So, $\left(x_1,y_1\right)=\left(-3,0\right)$ and $\left(x_2,y_2\right)=\left(9,0\right)$

So the coordinates of the mid-points of the line obtained by joining the foci of the ellipse, i.e., the coordinates of the center of the ellipse $\left(h,k\right)$ can be calculated as,

Coordinates the center of the ellipse $\left(h,k\right)=\left(\frac{-3+9}{2},\frac{0+0}{2}\right)$

$\Rightarrow$ $\left(h,k\right)=\left(\frac{6}{2},\frac{0}{2}\right)$

$\Rightarrow$ $\left(h,k\right)=\left(3,0\right)$

Step 5: From the standard equation of the ellipse

Since the center of the ellipse is $\left(3,0\right)$. And the values of $a$ and $b$ are $18$ and $12\sqrt{2}$ respectively.

And, the standard equation of the ellipse is,

$\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$

So, the equation of the given ellipse is,

$\frac{{\left(x-3\right)}^2}{{\left(18\right)}^2}+\frac{{\left(y-0\right)}^2}{{\left(12\sqrt{2}\right)}^2}=1$

So, the parametric equations of the given ellipse are,

$x-3=\left(18\right)\cos \theta$ $\left[\because x=acos\theta \right]$

$\Rightarrow$ $x=3+18\cos \theta$

And, $y-0=\left(12\sqrt{2}\right)\sin \theta$ $\left[\because y=bsin\theta \right]$

$\Rightarrow$ $y=12\sqrt{2}\sin \theta$

Thus, the parametric representation of a point of the ellipse is $\left(x,y\right)=\left(3+18\cos \theta ,12\sqrt{2}\sin \theta \right)$.

Hence, option (A) is the correct option.