Difficulty finding point A using two equations

Question

I have been having difficulty answering the following question:

The figure below shows a distribution between the origin, O, and the point B. The section from O to A is represented by the equation $y = x^2$, whilst the section from point A to point B is represented by the equation $3y = −20+ 20x −2x^2$. Find the coordinates of the point A.

Any help in terms of how to apply the correct methodology for this question and how to solve it would be greatly appreciated.

Answer 1

Let $ A=(x_0,y_0) $. A is the intersection point between the two curves, which means that $ y_0 = x_0^2 $ and that $ y_0 = \frac{1}{3} \big( -20 + 20x_0 - 2x_0^2 \big) $.

Let $ y_0 = y_0 $, i.e.

$$ x_0^2 = \frac{1}{3} \big( -20 + 20x - 2x^2 \big) $$

$$ 3x_0^2 = -20 + 20x_0 - 2x_0^2 $$

$$ 5x_0^2 - 20x_0 + 20 = 0 $$

$$ 5 (x_0^2 - 4x_0 + 4) = 0 $$

$$ 5(x_0-2)(x_0-2) = 0 $$

$$ x_0 = 2 $$

Plugging this into either of the two equations, you get: $ y_0 = x_0^2 = (2)^2 = 4 $. Therefore, $ A=(x_0,y_0) = (2,4). $

Answer 2

Plugging $y=x^2$ in the second equation, we get after simplification

$$x^2-4x+4=0$$ which gives the coordinates of the point $A : x=2$ and $y=4$.