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The sum of the squares of the distances ...

The sum of the squares of the distances of a moving point from two fixed points (a,0) and `(-a ,0)` is equal to a constant quantity `2c^2dot` Find the equation to its locus.

Text Solution

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Let `P(h,k)` be any position of the moveing point and let `A(a,0), B(-a,0)` be the given points. Then, we have `PA^2+PB^2=2c^2` (Given)
or `(h-a)^2+(k-0)^2+(h+a)^2+(k-0)^2=2c^2`
`2h^(2)2k^2+2a^2=2c^2`
or `h^2+k^2=c^2-a^2`
Hence, the equation to locus `(h,k)` is `x^2+y^2 =c^2-a^2`.
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Knowledge Check

  • If the square of distance of a moving point from the point (2,0) is equal to 4 units , then the locus of the moving point is a -

    A
    straight
    B
    circle
    C
    hyperbola
    D
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  • The relation between the locus of the points of equal distance from two fixed points is

    A
    equal bisector line
    B
    perpendicular line
    C
    parallel line
    D
    bisector of perpendicular
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