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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?

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

3 Answers 3

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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 $P_2(11/5,-4/\sqrt{5})$.

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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.

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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 $$\left(\frac{11}{5},-\frac{4}{\sqrt5}\right)$$

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