Find locus of the point moving such that sum of their distances from(3,0) and(-3,0) is less than 9 units.

Plz explain me the normal method as well as the other one in which we take PA+PB=2a and AB=2ae

Dear karhik,

 the locus will be all the points INSIDE the ellipse as the definition of ellipse says that it is the locus all the points whose sum of distances from the fixed points (focii) is constant and that is equal to 2 a where coordinates of focii are (ae,0) and (-ae,0) so if the distance is less than 9 

2a = 9    a=9/2

ae = 3    e = 2/3 

b² = a² (1 - e² )

b² = 45/4     a² = 81/4

ellipse is x² /(81/4)  +  y² /(45/4) = 1

alternate method   let the point be (h,k)    use distance formula {(h-3)² + (k-0)²}^1/2 + {(h+3)² + (k-0)²}^1/2 = 9

(h-3)² + (k-0)²}^1/2  = 9 - {(h+3)² + (k-0)²}^1/2

now square it  twice you will get the locus of the points and after getting the final equation we get the equation

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