A) \[{{({{x}^{2}}+{{y}^{2}})}^{3}}=4{{R}^{2}}{{x}^{2}}{{y}^{2}}\]
B) \[{{({{x}^{2}}+{{y}^{2}})}^{2}}=4R{{x}^{2}}{{y}^{2}}\]
C) \[({{x}^{2}}+{{y}^{2}})(x+y)={{R}^{2}}xy\]
D) \[{{({{x}^{2}}+{{y}^{2}})}^{2}}=4{{R}^{2}}{{x}^{2}}{{y}^{2}}\]
Correct Answer: A
Solution :
Let P(h, k) be the foot of the perpendicular from O to AB. \[\therefore \] Slope of \[AB=\frac{-h}{k}\] \[\therefore \]Equation of AB is \[hx+ky={{h}^{2}}+{{k}^{2}}\] \[\Rightarrow \]\[\frac{x}{\frac{{{h}^{2}}+{{k}^{2}}}{h}}+\frac{y}{\frac{{{h}^{2}}+{{k}^{2}}}{k}}=1\] So, coordinates of A and B are \[\left( \frac{{{h}^{2}}+{{k}^{2}}}{h},0 \right)\]and \[\left( 0,\frac{{{h}^{2}}+{{k}^{2}}}{k} \right)\]respectively. Now,\[AB=2R\Rightarrow {{(AB)}^{2}}=4{{R}^{2}}\] \[\Rightarrow \]\[\frac{{{({{h}^{2}}+{{k}^{2}})}^{2}}}{{{h}^{2}}}+\frac{{{({{h}^{2}}+{{k}^{2}})}^{2}}}{{{k}^{2}}}=4{{R}^{2}}\] \[\Rightarrow \]\[{{({{h}^{2}}+{{k}^{2}})}^{3}}=4{{R}^{2}}{{h}^{2}}{{k}^{2}}\] Thus, locus of P is \[{{({{x}^{2}}+{{y}^{2}})}^{3}}=4{{R}^{2}}{{x}^{2}}{{y}^{2}}.\]You need to login to perform this action.
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