Q57 / 150 · MHT CET 2007 · Mathematics

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

  • Correct answer
    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 the above

Solution

Distance between two foci, $2 a e=7+1=8$ $\therefore$ $$ a e=4 $$ $$ \Rightarrow $$ $a=8$ $$ \left(\because e=\frac{1}{2} \text { given }\right) $$ Now, $$ \begin{array}{l} b^{2}=a^{2}\left(1-e^{2}\right)=64\left(1-\frac{1}{4}\right) \\ b^{2}=48 \Rightarrow b=4 \sqrt{3} \end{array} $$ $\therefore$ Since, the centre of the ellipse is the mid point of the line joining two foci, therefore the coordinates of the centre are $(3,0)$. $\therefore$ Its equation is $$ \frac{(x-3)^{2}}{8^{2}}+\frac{(y-0)^{2}}{(4 \sqrt{3})^{2}}=1 $$ Hence, the parametric coordinates of a point on Eq. (i) are $(3+8 \cos \theta, 4 \sqrt{3} \sin \theta)$.

Hence, the correct answer is option: 'A'.

Relevant Questions