Question

The locus of the point represented by $x=1+4\cos \theta ,y=2+3\sin \theta$ is

• A. $9(x-1{)}^2-16(y-2{)}^2=1$
• B. $9(x-1{)}^2+16(y-2{)}^2=144$
• C. $16(x-1{)}^2-9(y-2{)}^2=1$
• D. $16(x-1{)}^2+9(y-2{)}^2=144$

Answer 1

The correct option is B $9(x-1{)}^2+16(y-2{)}^2=144$

$x=1+4\cos \theta$ and $y=2+3\sin \theta$

Hence

$\frac{x-1}{4}=\cos \theta$ and $\frac{y-2}{3}=\sin \theta$.

Therefore

${\cos }^2\theta +{\sin }^2\theta =1$

Or

${\left(\frac{x-1}{4}\right)}^2+{\left(\frac{y-2}{3}\right)}^2=1$

$\frac{{(x-1)}^2}{16}+\frac{{(y-2)}^2}{9}=1$

$\frac{9{(x-1)}^2+16{(y-2)}^2}{144}=1$

Or

$9{(x-1)}^2+16{(y-2)}^2=144$.