Q. $x = 4(1 + \cos{\theta)}\$ and $y = 3(1 + \sin{\theta)}$ are the parametric equations of
A) $\frac{(x - 3)^{2}}{9} + \frac{(y - 4)^{2}}{16} = 1$
B) $\frac{(x + 4)^{2}}{16} + \frac{(y + 3)^{2}}{9} = 1$
C) $\frac{(x - 4)^{2}}{16} - \frac{(y - 3)^{2}}{9} = 1$
D) $\frac{(x - 4)^{2}}{16} + \frac{(y - 3)^{2}}{9} = 1$
Explanation
(d)
Given, $\cos{\theta = \frac{x}{4} - 1}$ and $\sin{\theta = \frac{y}{3}} - 1$
$\therefore\ \cos^{2}\theta + \sin^{2}{\theta = 1}$
$\ \Rightarrow \ \ \ \left( \frac{x}{4} - 1 \right)^{2} + \left( \frac{y}{3} - 1 \right)^{2}\ = 1$
$\Rightarrow \ \frac{(x - 4)^{2}}{16} + \frac{(y - 3)^{2}}{9} = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.