# How do I find two intersection points from two parabolas?

Let's say I have two equations: $$4x+y^2=12$$ and $$x=y^2-1$$

I need to find the two intersection points of these parabolas so I can then calculate the enclosed area. I took a peek at what it would look like on Desmos.

So if I isolate for the $$y$$ variable for both equations, I managed to calculate the first intersection which is $$(\frac{11}{5}$$, $$\frac{4}{\sqrt5})$$.

How do I calculate the second intersection point? Do I just assume that since the function is a sideways parabola, then the other intersection point would just have a negative $$x$$ coordinate value?

• $y$ should be $\pm y$ Oct 28, 2020 at 5:55
• @MathLover is it because the {y} coordinate is a square root? Oct 28, 2020 at 6:13
• Yes that is correct. If you put $-y$ in both equations, it should give you the same value of $x$ as for $+y$. Oct 28, 2020 at 6:15

1)$$y^2=12-4x;$$

2)$$y^2=x+1;$$

Note: Both parabolas are symmetric about $$x-$$axis.

$$12-4x=x+1;$$

$$x=11/5.$$

Points of intersection:

1. $$y>0:$$ $$y=4/\sqrt{5}$$, e. g. $$P_1(11/5, 4/\sqrt{5})$$;

2. $$y<0:$$ By symmetry $$P_2(11/5,-4/\sqrt{5})$$.

Subtract one equation from the other, to eliminate y, and solve the resulting equation for x. Both solutions have the same x. Then plug the value of x into either parabola, and solve for y using the quadratic formula.

We have $$4(y^2-1)+y^2=12$$ or $$y^2=\frac{16}{5},$$ which gives $$y=\frac{4}{\sqrt5}$$ or $$y=-\frac{4}{\sqrt5}.$$ Also, $$x=\frac{16}{5}-1$$ or $$x=\frac{11}{5}$$ and we got two intersection points: $$\left(\frac{11}{5},\frac{4}{\sqrt5}\right)$$ and $$\left(\frac{11}{5},-\frac{4}{\sqrt5}\right)$$