# Solving two curves

Let two curves be:-

S1: $x^2/9+y^2/4=1$

S2: $y^2=2x$

Now, on solving these two ,by substituting $y^2$ as $2x$ from S2 in S1, I get two values viz. $x=3/2$ and $x=-6$.

But we can clearly see from the graph that the curves intersect at two distinct points in the first and fourth quadrant. So, should not the quadratic equation hence formed in $x$ rather be a whole square viz $(x-3/2)^2$ ?

• But why do we get it as a solution to the quadratic equations? Sep 28, 2016 at 15:30

When solving systems of nonlinear equations, naive substitution can introduce extraneous solutions. You really need to keep track of domains.

In this case, the equation $y^2=2x$ is only valid for $x\ge 0$. So when you substitute this into $\dfrac{x^2}9+\dfrac{y^2}4=1$, you have to take that into account. So if you don't want to have to plug your solutions back in at the end, you should have done your substitution with a note on the domain of $x$:

$$\frac{x^2}9+\frac{2x}{4}=1,\quad x\in[0,\infty)$$

Then you'd have gotten only one solution for $x$ -- and thus $2$ solutions for $(x,y)$.

Note that if you allow $x,y\in\Bbb C$, there actually are four solutions where two of them have $x=-6$. Can you find the $y$-values in that case?

You need to condition $x\geq 0$ since $y^2 = 2x$. Hence, $x=-6$ is not a solution.

• I understand that, but why is are we getting x=-6? Does it have any relation with the curves? Sep 28, 2016 at 15:36
• @ReeshabhRanjan When I solve equations like this, to keep it simple, I always keep the conditions straight. Here, S1 and S2 implies $x^2/9+2x/4=1$ ($\implies$, not $\iff$). To make them equivalent, you need the condition: S1 and S2 $\iff x^2/9+2x/4=1$ and $x\geq 0$ and $y^2 = 2x$. Sep 28, 2016 at 19:10

you have $y^2=3$ so

there are two possibilities

$y=\sqrt{3}$ or $y=-\sqrt{3}$ which gives 2 points.

In general, when you intersect two conics, you can have $4$ intersection points, that's why you see a ghost.

Imagine if you pinched the parabola and pulled it to the left through the other side of the ellipse. Then you would get $4$ intersections, and the other solution for $x$ would be relevant.

Two conics usually have four intersection points (by elimination of one variable you obtain a quartic equation). In this particular case, they are

$$\left(\frac32,3\right),\left(\frac32,-3\right),\left(-6,i\sqrt{12}\right),\left(-6,-i\sqrt{12}\right).$$

• What does the imaginary solution represent? Sep 28, 2016 at 15:52
• I have not studies much advance mathematics, just need an idea. Sep 28, 2016 at 15:52
• @ReeshabhRanjan: a short answer is difficult. It somehow means that the intersection does not lie in the usual plane, but in some "parallel world" that has four dimensions instead of two.
– user65203
Sep 28, 2016 at 16:01
• Oh means complex plane? Okay, that means that the other answer does not have any significance in the real plane. Sep 28, 2016 at 16:03
• @ReeshabhRanjan: yep, that's it. Just like complex roots of ordinary polynomials. We can compute them but they aren't "real".
– user65203
Sep 28, 2016 at 16:04