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 \[y=3(1+sn\theta )\] \[\Rightarrow \] \[\cos \theta =\frac{x}{4}-1\] and \[sin\theta =\frac{y}{3}-1\] We know \[{{\cos }^{2}}\theta +si{{n}^{2}}\theta =1\] \[\Rightarrow \] \[{{\left( \frac{x}{4}-1 \right)}^{2}}+{{\left( \frac{y}{3}-1 \right)}^{2}}=1\] \[\Rightarrow \] \[\frac{{{(x-4)}^{2}}}{16}+\frac{(y-3)}{9}=1\]
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