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# Locus of points- math question?

Given points F(0, √ 5) and F' (0, -√ 5) with PF+PF' = 8, find the equation of the locus point P(x,y).

Thanks!

### 1 Answer

- MadhukarLv 710 years agoFavorite Answer
This can easily be solved using the property of the ellipse and the equation of ellipse obtained. However, I shall attempt it from the first principles.

PF + PF' = 8

=> √[x^2 + (y - √5)^2] + √[x^2 + (y + √5)^2] = 8

=> √[x^2 + (y - √5)^2] = - √[x^2 + (y + √5)^2] + 8

Squarring,

=> x^2 + (y - √5)^2 = x^2 + (y + √5)^2 + 64 - 16√[x^2 + (y + √5)^2]

=> - 2√5 y = 2√5 y + 64 - 16√[x^2 + (y + √5)^2]

=> 4√[x^2 + (y + √5)^2] = √5 y + 16

=> 16 [x^2 + (y + √5)^2] = 5y^2 + 32√5 y + 256

=> 16x^2 + 16y^2 + 32√5 y + 80 = 5y^2 + 32√5 y + 256

=> 16x^2 + 11y^2 = 176

=> x^2/11 + y^2/16 = 1

which is the equation of the ellipse.

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Alternative method using the property of the ellipse:

For any point P on the ellipse, the sum of its distances from the foci is a constant = 2a, where 2a = length of the major axis. As the foce F and F' lie on the y-axis, the major axis is along the y-axis and is equal to 8

=> b = 4

Foci are F(0, be) and F' (0, -be)

=> be = √5

Also, a^2 = b^2(1 - e^2)

=> a^2 = b^2 - (be)^2

=> a^2 = 4^2 - 5

=> a^2 = 11

=> equation of the ellipse is

x^2/a^2 + y^2/b^2 = 1

=> x^2/11 + y^2/16 = 1.

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