Problem 01 - Equation of a curve

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

${d_1}^2 = (x + 3)^2 + y^2$
 

 

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

${d_2}^2 = (x - 3)^2 + y^2$
 

$d_1 + d_2 = 8$

$d_1 = 8 - d_2$

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

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

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

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

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

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

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

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

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

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

From the definition of ellipse, the given points (–3, 0) and (3, 0) are the foci and the sum 8 is the length of the major axis.
$c = 3$
 

$2a = 8$

$a = 4$

$a^2 = 16$
 

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

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

$b^2 = 7$
 

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

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

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

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