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Question

Find the equation of locus of a point such that, the difference of the squares of its distances from the point $$(5, 0)$$ and $$(2, 3)$$ is $$10$$.


Solution

$$\begin{array}{l} Let\, po{ { int } }\, be\left( { h,k } \right)  \\ dis\tan  ce\, from\, po{ { int } }\, \left( { 5,0 } \right) \, \, and\, \, \, \left( { 2,3 } \right) \, is \\ Dis\tan  ce=\sqrt { { { \left( { h-5 } \right)  }^{ 2 } }+{ { \left( { k-0 } \right)  }^{ 2 } } } .......\left( 1 \right)  \\ Dis\tan  ce=\sqrt { { { \left( { h-2 } \right)  }^{ 2 } }+{ { \left( { k-3 } \right)  }^{ 2 } } } .......\left( 2 \right)  \\ squaring\, both\, equation\, and\, substract\, \, \left( 2 \right) -\left( 1 \right)  \\ { \left[ { \sqrt { { { \left( { h-2 } \right)  }^{ 2 } }+{ { \left( { k-3 } \right)  }^{ 2 } } }  } \right] ^{ 2 } }-{ \left[ { \sqrt { { { \left( { h-5 } \right)  }^{ 2 } }+{ { \left( { k-0 } \right)  }^{ 2 } } }  } \right] ^{ 2 } }=10 \\ { \left( { h-2 } \right) ^{ 2 } }+{ \left( { k-3 } \right) ^{ 2 } }-\left[ { { { \left( { h-5 } \right)  }^{ 2 } }+{ k^{ 2 } } } \right] =10 \\ { h^{ 2 } }-2h+4+{ k^{ 2 } }-6k+9-\left[ { { h^{ 2 } }-10h+25+{ k^{ 2 } } } \right] =10 \\ { h^{ 2 } }-2h+4+{ k^{ 2 } }-6k+9-{ h^{ 2 } }+10h-25-{ k^{ 2 } }=10 \\ 8h-6k-25=10 \\ 8h-6k=35\, it\, is\, required\, locus \end{array}$$

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