conic with locus equation

Question

Find the equation of the locus of the point P(x, y) such that the sum of its distance to the points A(6,0) and B(-6, 0) is 18 units.

answer ((x-6)^2+y^2)^1/2 + ((x+6)^2+y^2)^1/2 = 18 5x^2-72x+9y^2-693=0

But answer given is x^2/81 + y^2/45 = 1

Answer 1

Usually, we denote the constant distance as $2a$ (in our case $2a = 18$) and the points $A,B$ are of the form $(c,0)$ and $(-c,0)$ (by the way, they are called the foci of the ellipse). Thus, we have: $$\begin{align} &(PA) + (PB) = 2a\\ &\sqrt{(x+c)^2 +y^2} = 2a - \sqrt{(x-c)^2 + y^2} &&{\text{square both sides and simplify}}\\ &a\sqrt{(x-c)^2 + y^2} = a^2 - cx &&{\text{square both sides}}\\ &a^2x^2 +a^2c^2-2a^2 cx +a^2y^2=a^4 +c^2x^2-2a^2cx\\ &(a^2-c^2)x^2+a^2y^2=a^2(a^2-c^2)\\ &\dfrac{x^2}{a^2}+\dfrac{y^2}{a^2-c^2}=1. \end{align}$$

Your initial thought is correct, but we have to do some algebraic manipulations in order to get to the given result.