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) \[\frac{{{(x-4)}^{2}}}{16}+\frac{{{(y-3)}^{2}}}{9}=1\]
Correct Answer: D
Solution :
We have \[x=4(1+\cos \theta )\]and\[y=3(1+\sin \theta )\] \[\Rightarrow \]\[\cos \theta =\frac{x}{4}-1\]and\[\sin \theta =\frac{y}{3}-1\] We know,\[co{{s}^{2}}\theta +si{{n}^{2}}\theta =1\] \[\Rightarrow \] \[{{\left( \frac{x}{4}-1 \right)}^{2}}+{{\left( \frac{y}{3}-1 \right)}^{2}}=1\] \[\Rightarrow \] \[\frac{{{(x-4)}^{4}}}{16}+\frac{{{(y-3)}^{2}}}{9}=1\]
You need to login to perform this action.
You will be redirected in
3 sec
