A) only y
B) only \[70{{\left( \frac{1-t}{1+t} \right)}^{4}}\]
C) both \[56{{\left( \frac{1+t}{1-t} \right)}^{3}}\] and y
D) neither \[56{{\left( \frac{1-t}{1+t} \right)}^{3}}\] nor y
Correct Answer: D
Solution :
\[\frac{{{x}^{2}}}{144}+\frac{{{y}^{2}}}{169}=1\] Applying \[\frac{{{x}^{2}}}{169}+\frac{{{y}^{2}}}{144}=1\] \[\frac{{{x}^{2}}}{12}+\frac{{{y}^{2}}}{13}=1\] Hence, \[{{x}^{2}}+{{y}^{2}}-2x+4y=0\]does not depend upon neither \[x\] nor \[y\].
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