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Chapter 32: Problem 1040

Find the equation for the set of points the sum of whose distances from \((4,0)\) and from \((-4,0)\) is 10 .

Short Answer

Expert verified
The equation for the set of points, the sum of whose distances from \((4,0)\) and from \((-4,0)\) is 10, is given by the ellipse equation: \[ x^2 + y^2 = 25 \]

Step by step solution

01

Identify the Foci Points

The foci (F1 and F2) are given as two points, namely \((4,0)\) and \((-4,0)\) respectively.
02

Define the sum of distances

We know that the sum of the distances from any point P on the ellipse to the two foci is constant. This is given as 10.
03

Formulate the Equation

If we take a point \(P(x,y)\) on the curve, according to the given conditions we would have: The distance between point P and F1 plus the distance between point P and F2 equals 10. Therefore, we can write this as: \[ \sqrt{(x-4)^2+y^2} + \sqrt{(x+4)^2+y^2} = 10 \] Squaring both sides to remove the square root: \[ [(x-4)^2+y^2] + 2\sqrt{[(x-4)^2+y^2][(x+4)^2+y^2]} +[(x+4)^2+y^2] = 100 \]
04

Simplify the equation

Simplify the above equation by cancelling out the equal terms and bring the terms involving the square root to one side: \[ 2\sqrt{[(x-4)^2+y^2][(x+4)^2+y^2]} = 100 - 2x^2 - y^2 \] Square both sides again to remove the square root: \[ 4[(x-4)^2+y^2][(x+4)^2+y^2] = (100 - 2x^2 - y^2)^2 \]
05

Obtain the final equation

Simplify the equation to remove complex terms. Since the square of the distances will always be positive, we can assume \(x^2 + y^2 = r^2\). The final equation of the ellipse would be: \[ x^2 + y^2 = 25 \] This is the final equation of the set of points the sum of whose distances from \((4,0)\) and from \((-4,0)\) is 10.

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