Solving the equations in three variables

Question

When I am dealing with some geometry problem in barycentric system I come across with the following equations

$\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}-xy\left(\frac{1}{a}+\frac{1}{b}\right)-yz\left(\frac{1}{b}+\frac{1}{c}\right)-xz\left(\frac{1}{a}+\frac{1}{c}\right)=0, c\left(b+c\right)x-acy+a\left(a+b\right)z=0 \ $

One is the equation of curve other is equation of line in barycentric coordinates

I attempted to solve these two equations by making y as a subject and substituting in the first equation but that doesn't give any solution,

Now my question is how to get x, y, z from these two equations or how to get x: y: z from these two equations in terms of a, b, c. here a, b, c are constants such as they are the sides of the triangle.

I got one point which is common for the both equations is x: y: z as a: 2s+b: c where 2s=a+b+c How to find other point which is common for both equations.

Thanks in advance

Answer 1

Clearing fractions, we can write

$$\begin{align} b c x^2 + c a y^2 + a b z^2 - ( a + b ) c x y - ( b + c ) a y z - ( c + a ) b z x &= 0 \\ [4pt] ( b + c ) c x - a c y + ( a + b ) a z &= 0 \end{align} \tag{1}$$

Now, we simply eliminate $v$ to get a relation we can write as $$(c x - a z) \left(\;c^2 (-a + b + c ) x - a^2 ( a + b - c ) z \;\right) = 0 \tag{2}$$ Thus, $$cx=az \quad\to\quad y = \frac{a + 2 b + c}{a} x \tag{3a}$$ or $$c^2 (-a + b + c ) x = a^2 ( a + b - c ) z \quad\to\quad y = \frac{b (a^2 + a b + b c + c^2)}{a^2 (a + b - c)}x \tag{3b}$$ so that

$$x:y:z \;=\; a:a+2b+c:c$$ or $$x:y:z \;=\; a^2(a+b-c):b \left(\;a(a+b) + c(b+c)\;\right):c^2(-a+b+c)$$