Intersection of a circle and a parabola

[rajeshmarndi]

We have a circle (x^2 + y^2=2) and a parabola (x^2=y).

We put x^2 = y in the circle equation and we get y^+y-2=0. We get two values of y as y=1 and y=-2.

Y=1 gives us two intersection point i.e (1,1) and (-1,1). But y=-2 neither it lie on the circle nor on the parabola. The discriminant of the quadratic equation y^+y-2=0 is positive but yet, why do we get the other solution(y=-2) in complex form?

Thank you.

[Moderator's note: Moved from a technical forum and thus no template.]

 

[Infrared]

There there are in fact (complex) solutions with $y=-2$. The points $(i\sqrt{2},-2)$ and $(-i\sqrt{2},-2)$ lie on both curves.

There's nothing unusual about having some real and some non-real solutions. For a simpler example, the function $y=x(x^2+1)$ has one real root, and two non-real roots.

 

[rajeshmarndi]

That mean, even if the discriminant of the quadratic equation is positive, we can still get solution in complex form.

 

[Mark44]

There there are in fact (complex) solutions with $y=-2$.

I doubt very much that complex solutions are relevant in this problem.

The given equations are $x^2 + y^2 = 2$ and $y = x^2$. From the second equation, y must be nonnegative since it's the square of a real number. The equation obtained after replacing $x^2$ in the circle equation resulted in the quadratic equation $y^2 + y - 2 = 0$ whose solutions were y = 1 and y = -2.

Since y =-2 is not a solution of the equation $y = x^2$, there won't be any points common to both curves with this y-value.

 

[haruspex]

I doubt very much that complex solutions are relevant in this problem.

They are not relevant to the problem to be solved, but they are relevant to the OP's question: where do these complex solutions come from?

If we allow complex values of x then we can visualise them by plotting the imaginary part on the z axis. For the circle, we get a rectangular hyperbola in the YZ plane, meeting the circle at (0, ±√2, 0). For the parabola, we get a parabola in the negative y half of the YZ plane, meeting the original parabola orthogonally at the origin.
The additional solutions are where these curves intersect.

 

[ehild]

We have a circle (x^2 + y^2=2) and a parabola (x^2=y).

We put x^2 = y in the circle equation and we get y^+y-2=0. We get two values of y as y=1 and y=-2.

Y=1 gives us two intersection point i.e (1,1) and (-1,1). But y=-2 neither it lie on the circle nor on the parabola. The discriminant of the quadratic equation y^+y-2=0 is positive but yet, why do we get the other solution(y=-2) in complex form?

When you replaced x^2 by y you got a quadratic equation for y with two real roots. f(y) =y^2+y-2 is a parabola and it intersects the y-axis in two points. This problem is different from the original one. In the original problem, negative value was not allowed for y, as y was the square of a real number. So y=-2 had to be discarded.
If you replace y by x^2 you got a fourth-order equation in x: x^4+x^2-2=0. That equation has two real roots and two complex (imaginary) ones
When you give the solution of the original problem you have two check which roots are relevant you will discard the imaginary x values.