Point of intersection between a parabola and a circle

[sooyong94]

Homework Statement

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.

Homework Equations

Equation of a circle, parametric equations

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?

 

[SteamKing]

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

 

[sooyong94]

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

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 $r<-\frac{\sqrt{7}}{2}$ ?

 

[SteamKing]

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

 

[sooyong94]

Well, the solution checks out!

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

 

[haruspex]

Well, the solution checks out!

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

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.

 

[sooyong94]

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.

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

 

[SammyS]

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

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?

 

[sooyong94]

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?

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

 

[haruspex]

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

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.

 

[sooyong94]

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.

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

 

[SammyS]

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

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:

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

 

[sooyong94]

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:

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

But why r>2?

 

[SammyS]

But why r>2?

Look at that graph.

 

[sooyong94]

Look at that graph.

Like this?

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

 

[SammyS]

Yes .

 

[haruspex]

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

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?

 

[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

 

[haruspex]

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

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.

 

[sooyong94]

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?

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

 

[haruspex]

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

Then see my post #19.

 

[sooyong94]

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.

So what should I do next? Use a substitution?

 

[SammyS]

So what should I do next? Use a substitution?

Yes, that's one way to start.

What substitution do you have in mind? One possibility stands out.

 

[sooyong94]

$u=t^{2}$ ? I did tried that. :(

 

[SammyS]

$u=t^{2}$ ? I did tried that. :(

So, that gives:

$u^{2} -3u +4-r^{2}=0\$ ,​

Right?

Putting t² back in for u gives

$\displaystyle \ t^2=\frac{3}{2} \pm \sqrt{r^2-(7/2)\,}\$ .​

Now consider all the possibilities.

 

[sooyong94]

So, that gives:

$u^{2} -3u +4-r^{2}=0\$ ,​

Right?

Putting t² back in for u gives

$\displaystyle \ t^2=\frac{3}{2} \pm \sqrt{r^2-(7/2)\,}\$ .​

Now consider all the possibilities.

What possibilities? Do you mean the number of solutions?

 

[SammyS]

What possibilities? Do you mean the number of solutions?

Number and what kind -- real or complex (with non-zero) imaginary part, and multiplicity.

 

[sooyong94]

Number and what kind -- real or complex (with non-zero) imaginary part, and multiplicity.

So I need to find the set values for which it only has one solution right?