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# Equation of the locus point?

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

I don't even know where to begin with this one.

### 2 Answers

- JLv 710 years agoFavorite Answer
PF = distance (x,y) to (0, √5) is √[(x-0)^2 + (y-√5)^2] = √[x^2 + (y-√5)^2] , by Pyth thm (distance formula)

PF' = distance (x,y) to (0,-√5) is √[(x-0)^2 + (y+√5)^2] = √[x^2 + (y+√5)^2].

The equation PF + PF' = 8, after tranposing PF' is

√[x^2 + (y-√5)^2] = 8 - √[x^2 + (y+√5)^2], now square both sides

x^2 + (y-√5)^2 = 64 - 16√[x^2 + (y+√5)^2] + x^2 + (y+√5)^2, expand squares and simplify

4√5 y +64 = 16√[x^2+y^2+2y√5+5], divide by 4 and square again

5y^2+32y√5+256 = 16[x^2+y^2+2y√5+5], simplify

16x^2 + 11y^2 = 176, divide by 176

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

Note: this is an ellipse

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- desmaraisLv 43 years ago
i might like to grant this a crack, yet regrettably, i'm previous due as that is. the way i might attitude it would be to construct a parabola making use of the concentration-directorix shape. you need to use a vector equation for the directorix. substitute in the coordinates of the factors to get some circumstances for the variables in the parabola's shape. with a bit of luck, those circumstances will spit out an equation for the locus.

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