Solving two curves

Question

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 $(x-3/2)^2$ ?

Answer 1

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

Answer 2

you have $y^2=3$ so

there are two possibilities

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

Answer 3

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.

Answer 4

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?

Answer 5

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

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