A) 8
B) 6
C) 10
D) 12
Correct Answer: C
Solution :
Equation of the curve is \[\frac{{{x}^{2}}}{{{5}^{2}}}+\frac{{{y}^{2}}}{{{4}^{2}}}=1\] Þ\[-5\le x\le 5,\,\,-4\le y\le 4\] \[P{{F}_{1}}+P{{F}_{2}}=\sqrt{[{{(x-3)}^{2}}+{{y}^{2}}]}+\sqrt{[{{(x+3)}^{2}}+{{y}^{2}}]}\] \[=\sqrt{{{(x-3)}^{2}}+\frac{400-16{{x}^{2}}}{25}}+\sqrt{{{(x+3)}^{2}}+\frac{400-16{{x}^{2}}}{25}}\] \[=\frac{1}{5}\left\{ \sqrt{(9{{x}^{2}}+625-150x)}+\sqrt{(9{{x}^{2}}+625+150x)} \right\}\] \[=\frac{1}{5}\left\{ \sqrt{{{(3x-25)}^{2}}}+\sqrt{{{(3x+25)}^{2}}} \right\}=\frac{1}{5}\left\{ 25-3x+3x+25 \right\}\] \[=10\], \[(\because 25-3x>0,\,25+3x>0\])You need to login to perform this action.
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