The sum of the squares of the distances of a | vedclass

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Question

(A) Let the moving point be $P(x, y)$.Given fixed points are $A(a, 0)$ and $B(-a, 0)$.The square of the distance $AP$ is $AP^2 = (x-a)^2 + (y-0)^2 = (

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$x^2+y^2=c^2-a^2$

Vedclass · Answer

(A) Let the moving point be $P(x, y)$.Given fixed points are $A(a, 0)$ and $B(-a, 0)$.The square of the distance $AP$ is $AP^2 = (x-a)^2 + (y-0)^2 = (x-a)^2 + y^2$.The square of the distance $BP$ is $BP^2 = (x+a)^2 + (y-0)^2 = (x+a)^2 + y^2$.According to the problem,the sum of these squares is $2c^2$:$AP^2 + BP^2 = 2c^2$$(x-a)^2 + y^2 + (x+a)^2 + y^2 = 2c^2$$(x^2 - 2ax + a^2) + y^2 + (x^2 + 2ax + a^2) + y^2 = 2c^2$$2x^2 + 2y^2 + 2a^2 = 2c^2$Dividing by $2$,we get:$x^2 + y^2 + a^2 = c^2$$x^2 + y^2 = c^2 - a^2$Thus,the equation of the locus is $x^2 + y^2 = c^2 - a^2$.

The sum of the squares of the distances of a moving point from two fixed points $A(a, 0)$ and $B(-a, 0)$ is equal to a constant $2c^2$. Then,the equation of its locus is:

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