# Point of intersection between a parabola and a circle

1. May 13, 2014

### sooyong94

1. The problem statement, all variables and given/known data
Sketch the curve C defined parametrically by
$x=t^{2} -2, y=t$

Write down the Cartesian equation of the circle with center as the origin and radius $r$. Show that this circle meets the curve C at points whose parameter $t$ satisfies the equation
$t^{4} -3t^{2} +4-r^{2}=0$

(a) In the case $r=2\sqrt{2}$, find the coordinates of the two points of intersection of the curve and the circle

(b) Find the range of values of $r$ for which the curve and the circle have exactly two points in common.

2. Relevant equations
Equation of a circle, parametric equations

3. The attempt at a solution
$x=t^{2} -2, y=t$
$x=y^{2} -2$
$y^{2}=x+2$
The equation represent a parabola.

The equation of a circle with origin as the center and radius $r$ is given by
$x^{2}+y^{2} =r^{2}$

Substituting $x=t^{2} -2$ and $y=t$, and simplifying it gives
$t^{4} -3t^{2} +4-r^{2}=0$

(a) Taking $r=2\sqrt{2}$, and solving it for t, I have t=+-2, and hence x=2 and y=2, as well as x=2 and y=-2.

(b) Now I'm stuck at this part. I know I can use quadratic discriminant for quadratic equations, but this is a quartic equation. How do I work this out?

2. May 13, 2014

### SteamKing

Staff Emeritus
It's a quartic equation in t, but a simple substitution can turn your quartic in t into a quadratic in another variable.

3. May 13, 2014

### sooyong94

I know since the circle cuts the parabola at two points, therefore $b^{2} -4ac >0$?

Now, if I let $u=t^{2}$, then using the quadratic discriminant, then $r>\frac{\sqrt{7}}{2}$ or $r<-\frac{\sqrt{7}}{2}$ ?

4. May 13, 2014

### SteamKing

Staff Emeritus
You can always draw a sketch and see if those values of r satisfy the original requirements for the circle and the parabola.

5. May 13, 2014

### sooyong94

Well, the solution checks out!

But why this works when r=sqrt(7) /2 but not greater?

Last edited: May 13, 2014
6. May 14, 2014

### haruspex

I'm a bit confused too.
You posted that $r>\frac{\sqrt{7}}{2}$ or $r<-\frac{\sqrt{7}}{2}$. r is a radius; r and -r yield the same circle, so it's usual to take only r ≥ 0.
Looking at the graph, what will happen as you increase r? The OP asks for those r where there are exactly two points of intersection.

7. May 14, 2014

### sooyong94

When I increase the value of r, it seems like there are four solutions...

8. May 14, 2014

### SammyS

Staff Emeritus
How much can you increase r and still have 4 solutions?

What if you continue to increase r beyond that? Can you then get fewer solutions?

9. May 14, 2014

### sooyong94

Well, yeah, but I found the answer should be r>2...

10. May 14, 2014

### haruspex

Not sure what you are saying there.
Are you saying you have been told the answer is r > 2, but you don't understand why, or that by following SammyS's advice you now get the answer r > 2? Either way, that is not the complete answer since you already correctly found one value less than that.

11. May 15, 2014

### sooyong94

I was meant that the book's answer tells me that the correct answer is r>2... :(

12. May 15, 2014

### SammyS

Staff Emeritus
Yes when posting an answer from the book, it reduces confusion to state the source of the answer.

haruspex's comment means that the complete answer to part (b) is:

$\displaystyle \left\{ r: r=\frac{\sqrt{7}}{2} \text{ or } r>2 \right\}\ .$

13. May 15, 2014

### sooyong94

But why r>2?

14. May 15, 2014

### SammyS

Staff Emeritus
Look at that graph.

15. May 15, 2014

### sooyong94

Like this?

However I don't know how to get the inequality r>2.

16. May 15, 2014

### SammyS

Staff Emeritus
Yes .

17. May 16, 2014

### haruspex

Do you mean that you don't know how to get it from the graph? How many points of contact when r=2. How many when r is a bit > 2? Will the number change again as r increases to infinity?

Or do you mean that you can see it from the graph but want to derive it algebraically?

18. May 16, 2014

### ehild

Solve the equation $t^{4} -3t^{2} +4-r^{2}=0$ for t2 and figure out when one of the roots is negative.

ehild

19. May 16, 2014

### haruspex

I don't think that's quite precise enough.
A value of r can produce one or two solutions for t2.
Every positive real value for t2 will yield two real and distinct solutions for t.
Complex values for t2 yield no real solutions for t.
Negative real values for t2 yield no real solutions for t.
t2 = 0 yields one real solution for t.
It follows that we need r to produce exactly one positive real value for t2 and no zero values. There are two distinct ways it can do this: by producing a repeated real root of the equation for t2, or by producing one positive real root and one negative one.

20. May 16, 2014

### sooyong94

I can see it from the graph but I want to solve it algebraically... :P