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 x^{2}/a^{2} + y^{2}/a^{2}(1-e^{2}) = 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} + k^{2}]^{1/2} = 2a - [(h + ae)^{2} + k^{2}]

on^{ }squaring both sides,

h^{2} + a^{2}e^{2} -2aeh + k^{2} = 4a^{2} + h^{2} + a^{2}e^{2} + 2aeh +k^{2} - 4a[(h + ae)^{2} + k^{2}]^{1/2}

4aeh + 4a^{2} = 4a[(h + ae)^{2} + k^{2}]^{1/2}

eh +a = [(h + ae)^{2} + k^{2}]^{1/2}

again, on squaring both sides,

a^{2} + e^{2}h^{2} + 2aeh = h^{2} + a^{2}e^{2} +2aeh + k^{2}

a^{2} + e^{2}h^{2} = h^{2} + a^{2}e^{2} + k^{2}

h^{2} - e^{2}h^{2} + k^{2} = a^{2} - a^{2}e^{2}

h^{2}(1 - e^{2}) + k^{2} = a^{2}(1 - e^{2})

Divide both sides by a^{2}(1 - e^{2}),

h^{2}/a^{2} + k^{2}/a^{2}(1 - e^{2}) = 1

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

x^{2}/a^{2} + y^{2}/a^{2}(1 - e^{2}) = 1

My answer is obviously 100% correct. So, give me a thumps up.

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