# 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?

May 1, 2016

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

#### Explanation:

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

$\sqrt{{\left(x - 3\right)}^{2} + {\left(y - 1\right)}^{2}} + \sqrt{{\left(x + 1\right)}^{2} + {\left(y - 1\right)}^{2}} = 6$ or

$\sqrt{{\left(x - 3\right)}^{2} + {\left(y - 1\right)}^{2}} = 6 - \sqrt{{\left(x + 1\right)}^{2} + {\left(y - 1\right)}^{2}}$

Squaring each side, we get

${\left(x - 3\right)}^{2} + {\left(y - 1\right)}^{2} = 36 + {\left(x + 1\right)}^{2} + {\left(y - 1\right)}^{2} - 12 \sqrt{{\left(x + 1\right)}^{2} + {\left(y - 1\right)}^{2}}$ or

${x}^{2} - 6 x + 9 + {y}^{2} - 2 y + 1 = 36 + {x}^{2} + 2 x + 1 + {y}^{2} - 2 y + 1 - 12 \sqrt{{\left(x + 1\right)}^{2} + {\left(y - 1\right)}^{2}}$ or

$- 6 x + 9 = 36 + 2 x + 1 - 12 \sqrt{{\left(x + 1\right)}^{2} + {\left(y - 1\right)}^{2}}$ or

$12 \sqrt{{\left(x + 1\right)}^{2} + {\left(y - 1\right)}^{2}} = 36 + 2 x + 1 + 6 x - 9 = 8 x + 28$ or

$3 \sqrt{{\left(x + 1\right)}^{2} + {\left(y - 1\right)}^{2}} = 2 x + 7$ and squaring again

$9 \left({\left(x + 1\right)}^{2} + {\left(y - 1\right)}^{2}\right) = 4 {x}^{2} + 28 x + 49$ or

$9 \left({x}^{2} + 2 x + 1 + {y}^{2} - 2 y + 1\right) = 4 {x}^{2} + 28 x + 49$ or

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

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]}