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The locus of a points p which moves such that 2PA = 3PB, where \( \Large A \left(0,\ 0\right)\ and\ B \left(4,\ -3\right) \) are points is:
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A) \( \Large 5x^{2}-5y^{2}-72x+54y+225=0 \) |
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B) \( \Large 5x^{2}+5y^{2}-72x+54y+225=0 \) |
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C) \( \Large 5x^{2}+5y^{2}+72x-54y+225=0 \) |
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D) \( \Large 5x^{2}+5y^{2}-72x-54y+225=0 \) |
Correct Answer:
| B) \( \Large 5x^{2}+5y^{2}-72x+54y+225=0 \) |
Description for Correct answer:
Let \( \Large P \left(h,\ k\right) \) be the required point then
\( \Large 4PA^{2} = 9PB^{2} \)
=> \( \Large 4 \left(h^{2}+k^{2}\right)=9 \left(h-4\right)^{2}+9 \left(k+3\right)^{2} \)
=> \( \Large 4h^{2}+4k^{2}=9 \left(h^{2}+16-8h\right) + 9 \left(k^{2}+9+6k\right) \)
=> \( \Large 5h^{2}+5k^{2}-72h+54k+225=0 \)
Therefore, Required locus of \( \Large P \left(h,\ k\right)\ is \)
\( \Large 5x^{2}+5y^{2}-72x+54y+225=0 \)
Part of solved Rectangular and Cartesian products questions and answers : >> Elementary Mathematics >> Rectangular and Cartesian products
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