Question

The equation of the curve whose parametric equations are $x=1+4\cos \theta ,y=2+3\sin \theta ,\theta \in R,$ is

• A. $16x^2+9y^2-64x-18y-71=0$
• B. $9x^2+16y^2-18x-64y-71=0$
• C. $9x^2+16y^2-18x-64y+71=0$
• D. $16x^2+9y^2-64x-18y+71=0$

Answer 1

The correct option is B $9x^2+16y^2-18x-64y-71=0$

We have $x=1+4\cos \theta ,y=2+3\sin \theta$.
$x=1+4\cos \theta$
$\Rightarrow \frac{x-1}{4}=\cos \theta \cdots \left(1\right)$
and
$y=2+3\sin \theta$
$\Rightarrow \frac{y-2}{3}=\sin \theta \cdots \left(2\right)$
Squaring and adding equation $\left(1\right)$ and $\left(2\right)$, we get
$\therefore \frac{(x-1{)}^2}{4^2}+\frac{(y-2{)}^2}{3^2}={\sin }^2\theta +{\cos }^2\theta$
$\Rightarrow \frac{(x-1{)}^2}{4^2}+\frac{(y-2{)}^2}{3^2}=1$
$\Rightarrow 9(x-1{)}^2+16(y-2{)}^2=144$
$\Rightarrow 9x^2+16y^2-18x-64y-71=0$