How do you find the equation of the following conic section and identify it given: All points such that the sum of the distance to the points (3,1) and (-1,1) equals 6?

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Answer 1 · 2016-05-01T06:39:20

Equation is $5x^2+9y^2-10x-18y-31=0$ and is of an ellipse.

Let the point on the locus be $(x,y)$ , sum of whose distances from $(3,1)$ and $(-1,1)$ is $6$ , hence

$sqrt((x-3)^2+(y-1)^2)+sqrt((x+1)^2+(y-1)^2)=6$ or

$sqrt((x-3)^2+(y-1)^2)=6-sqrt((x+1)^2+(y-1)^2)$

Squaring each side, we get

$(x-3)^2+(y-1)^2=36+(x+1)^2+(y-1)^2-12sqrt((x+1)^2+(y-1)^2)$ or

$x^2-6x+9+y^2-2y+1=36+x^2+2x+1+y^2-2y+1-12sqrt((x+1)^2+(y-1)^2)$ or

$-6x+9=36+2x+1-12sqrt((x+1)^2+(y-1)^2)$ or

$12sqrt((x+1)^2+(y-1)^2)=36+2x+1+6x-9=8x+28$ or

$3sqrt((x+1)^2+(y-1)^2)=2x+7$ and squaring again

$9((x+1)^2+(y-1)^2)=4x^2+28x+49$ or

$9(x^2+2x+1+y^2-2y+1)=4x^2+28x+49$ or

$5x^2+9y^2-10x-18y-31=0$

As the coefficient of $x^2$ and $y^2$ are positive but different, it is an ellipse.

graph{5x^2+9y^2-10x-18y-31=0 [-10, 6, -5, 5]}