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Question: The locus of the moving point P, such that 2PA = 3PB where A is (0,0) and B is (4,–3), is....

The locus of the moving point P, such that 2PA = 3PB where A is (0,0) and B is (4,–3), is.

A

5x25y272x+54y+225=05x^{2} - 5y^{2} - 72x + 54y + 225 = 0

B

5x25y2+72x+54y+225=05x^{2} - 5y^{2} + 72x + 54y + 225 = 0

C

5x2+5y2+72x+54y+225=05x^{2} + 5y^{2} + 72x + 54y + 225 = 0

D

5x2+5y272x+54y+225=05x^{2} + 5y^{2} - 72x + 54y + 225 = 0

Answer

5x2+5y272x+54y+225=05x^{2} + 5y^{2} - 72x + 54y + 225 = 0

Explanation

Solution

Let P(h,k)P ( h , k ) Given 2PA=3PB4PA2=9PB22 P A = 3 P B \Rightarrow 4 P A ^ { 2 } = 9 P B ^ { 2 }

4(h2+k2)=9[(h4)2+(k+3)2]\Rightarrow 4 \left( h ^ { 2 } + k ^ { 2 } \right) = 9 \left[ ( h - 4 ) ^ { 2 } + ( k + 3 ) ^ { 2 } \right]

5h2+5k272h+54k+225=0\Rightarrow 5 h ^ { 2 } + 5 k ^ { 2 } - 72 h + 54 k + 225 = 0

Hence the locus of point P is given by

5x2+5y272x+54y+225=05 x ^ { 2 } + 5 y ^ { 2 } - 72 x + 54 y + 225 = 0.