**Problem 01**

Determine the equation of the curve such that the sum of the distances of any point of the curve from two points whose coordinates are (–3, 0) and (3, 0) is always equal to 8.

**Solution 01**

${d_1}^2 = (x + 3)^2 + y^2$

$d_2 = \sqrt{(x - 3)^2 + y^2}$

${d_2}^2 = (x - 3)^2 + y^2$

$d_1 + d_2 = 8$

$d_1 = 8 - d_2$

${d_1}^2 = (8 - d_2)^2$

${d_1}^2 = 64 - 16d_2 + {d_2}^2$

$(x + 3)^2 + y^2 = 64 - 16\sqrt{(x - 3)^2 + y^2} + [ \, (x - 3)^2 + y^2 \, ]$

$(x^2 + 6x + 9) + y^2 = 64 - 16\sqrt{(x - 3)^2 + y^2} + (x^2 - 6x + 9) + y^2$

$12x - 64 = -16\sqrt{(x - 3)^2 + y^2}$

$3x - 16 = -4\sqrt{(x - 3)^2 + y^2}$

$(3x - 16)^2 = 16[ \, (x - 3)^2 + y^2 \, ]$

$9x^2 - 96x + 256 = 16[ \, (x^2 - 6x + 9) + y^2 \, ]$

$9x^2 - 96x + 256 = (16x^2 - 96x + 144) + 16y^2$

$7x^2 + 16y^2 - 112 = 0$ ← ellipse *answer*

**Another Solution**

$c = 3$

$2a = 8$

$a = 4$

$a^2 = 16$

$b^2 + c^2 = a^2$

$b^2 + 3^2 = 4^2$

$b^2 = 7$

$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$

$\dfrac{x^2}{16} + \dfrac{y^2}{7} = 1$

$7x^2 + 16y^2 = 16(7)$

$7x^2 + 16y^2 - 112 = 0$ *answer*

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