Ellipse
6.0 Parametric Co-ordinates
6.0 Parametric Co-ordinates
The equation of ellipse is $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$. Clearly, $x = a\cos \theta $ and $y = b\sin \theta $ satisfy the equation of ellipse for all real values of $\theta $. Hence, the parametric coordinates of ellipse are $(a\cos \theta ,b\sin \theta )$ where $\theta (0 \leqslant \theta < 2\pi )$ is the parameter and also called the eccentric angle.
Question 6. Find the eccentric angle on the point of the ellipse ${x^2} + 3{y^2} = 6$ at the distance of $2$ units from the centre of ellipse.
Solution: The equation of ellipse can be written as $$\frac{{{x^2}}}{6} + \frac{{{y^2}}}{2} = 1$$
On comparing it with the standard equation of ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, we get ${a^2} = 6$ and ${b^2} = 2$. Therefore, the parametric coordinates of a point $P$ on the ellipse can be written as $(\sqrt 6 \cos \theta ,\sqrt 2 \sin \theta )$. As given in the question, the distance of point $P$ is $2$ units from the centre of ellipse i.e., $(0,0)$. Using Distance formulae, we get $$\begin{equation} \begin{aligned} 6{\cos ^2}\theta + 2{\sin ^2}\theta = 4 \\ 4{\cos ^2}\theta + 2 = 4 \\ 4{\cos ^2}\theta = 2 \\ \cos \theta = \pm \frac{1}{{\sqrt 2 }} \\\end{aligned} \end{equation} $$
If $\cos \theta = \frac{1}{{\sqrt 2 }}$, $\theta = - \frac{\pi }{4}$ or $\frac{{7\pi }}{4}$.
If $\cos \theta = - \frac{1}{{\sqrt 2 }}$, $\theta = \frac{{3\pi }}{4}$ or $\frac{{5\pi }}{4}$.
