A point moves such that the sum of its distances from the point (ae,0) and (-ae,0) is 2a. Show that the locus of this point is x2/a2 + y2/a2(1-e2) = 1

let the point be (h,k). Then, By using distance formula.

[(h - ae)2 + (k - 0)2]1/2 + [(h+ae)2 + (k-0)2]1/2 = 2a

[(h - ae)2 + k2]1/2 = 2a - [(h + ae)2 + k2]

on squaring both sides,

h2 + a2e2 -2aeh + k2 = 4a2 + h2 + a2e2 + 2aeh +k2 - 4a[(h + ae)2 + k2]1/2

4aeh + 4a2 = 4a[(h + ae)2 + k2]1/2

eh +a = [(h + ae)2 + k2]1/2

again, on squaring both sides,

a2 + e2h2 + 2aeh = h2 + a2e2 +2aeh + k2

a2 + e2h2 = h2 + a2e2 + k2

h2 - e2h2 + k2 = a2 - a2e2

h2(1 - e2) + k2 = a2(1 - e2)

Divide both sides by a2(1 - e2),

h2/a2 + k2/a2(1 - e2) = 1

To get the locus of this point (h, k), replace (h, k) by (x, y).

x2/a2 + y2/a2(1 - e2) = 1

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