Find the locus of a point which moves so that the sum of its distances from $\left(3,0\right)$ and $\left(-3,0\right)$ is $9$.
Let $P \left(x, y\right)$ be the variable point.
Let $F_1 = \left(3,0\right)$ and $F_2 = \left(-3,0\right)$
Then as per question $\;$ $PF_1 + PF_2 = 9$
i.e. $\;$ $\sqrt{\left(x - 3\right)^2 + \left(y - 0\right)^2} + \sqrt{\left(x + 3\right)^2 + \left(y - 0\right)^2} = 9$
i.e. $\;$ $\sqrt{x^2 - 6x + 9 + y^2} - 9 = \sqrt{x^2 + 6x + 9 + y^2}$ $\;\;\; \cdots \; (1)$
Squaring both sides of equation $(1)$ gives
$x^2 - 6x + 9 + y^2 + 81 - 18 \sqrt{x^2 - 6x + 9 + y^2} = x^2 + 6 x + 9 + y^2$
i.e. $\;$ $- 12 x + 81 = 18 \sqrt{x^2 - 6x + 9 + y^2}$
i.e. $\;$ $-4 x + 27 = 6 \sqrt{x^2 - 6x + 9 + y^2}$ $\;\;\; \cdots \; (2)$
Squaring both sides of equation $(2)$ gives
$16x^2 + 729 - 216 x = 36 \left(x^2 - 6x + 9 + y^2\right)$
i.e. $\;$ $20 x^2 + 36 y^2 = 405$
i.e. $\;$ $\dfrac{x^2}{405 / 20} + \dfrac{y^2}{405 / 36} = 1$
i.e. $\;$ $\dfrac{x^2}{81 / 4} + \dfrac{y^2}{45 / 4} = 1$ $\;\;\; \cdots \; (3)$
Equation $(3)$ gives the required equation of locus.