A) Ellipse
B) Parabola
C) Hyperbola
D) None of these
Correct Answer: A
Solution :
Let point P \[({{x}_{1}},{{y}_{1}})\] So,\[\sqrt{{{({{x}_{1}}+2)}^{2}}+y_{1}^{2}}=\frac{2}{3}\left( {{x}_{1}}+\frac{9}{2} \right)\] Þ \[{{({{x}_{1}}+2)}^{2}}+y_{1}^{2}=\frac{4}{9}{{\left( {{x}_{1}}+\frac{9}{2} \right)}^{2}}\] Þ \[9[x_{1}^{2}+y_{1}^{2}+4{{x}_{1}}+4]=4\left( x_{1}^{2}+\frac{81}{4}+9{{x}_{1}} \right)\] Þ \[5x_{1}^{2}+9y_{1}^{2}=45\]Þ\[\frac{x_{1}^{2}}{9}+\frac{y_{1}^{2}}{5}=1\], Locus of \[({{x}_{1}},\,{{y}_{1}})\] is\[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{5}=1\], which is equation of an ellipse.
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