Question

$x=4(1+\cos \theta )$ 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. $\frac{{(x-4)}^2}{16}+\frac{{(y-3)}^2}{9}=1$

Answer 1

The correct option is D $\frac{{(x-4)}^2}{16}+\frac{{(y-3)}^2}{9}=1$

Given, $x=4(1+\cos \theta )$ and $y=3(1+\sin \theta )$
$\Rightarrow \cos \theta =\frac{x}{4}-1$
and $\sin \theta =\frac{y}{3}-1$
We know that
${\cos }^2\theta +{\sin }^2\theta =1$
$\Rightarrow {(\frac{x}{4}-1)}^2+{(\frac{y}{3}-1)}^2=1$
$\Rightarrow \frac{{(x-4)}^2}{16}+\frac{{(y-3)}^2}{9}=1$