# How did they get these 3 boundaries?

unscientific

## Homework Statement

The question is attached in the picture.

## The Attempt at a Solution

From the question I deduced that there are 3 boundaries: red, green and blue.

Blue Boundary

This one is the easiest:

That part of the curve of y2 = 4a(a-x), which corresponds to u = a.

Red Boundary

This corresponds to x = 0, y > 0.
When x = 0,

y2 = 4u2 = 4v2, which corresponds to u = v.

Blue Boundary

This happens when y = 0, x > 0.

When y = 0,

u = 0 or u = x. ------ (1)

v = 0 or v = -x.

Not sure what is going on here...

To try and vizualize the boundaries in the u-v (u,v) plane, I tried to find the intersection between each boundary.

Blue-red Intersection

u = a AND u = v

Therefore, this corresponds to the point (a, a)

I've yet to find the red-green intersection and the blue-green intersection..

#### Attachments

tt2348
I have a better idea..
Instead of going through and trying to formulate each of these piece by piece, try expressing x=x(u,v) and y=y(u,v)
For instance
$y^{2}=4u(u-x)=4v(v+x)\Rightarrow 4(u^2-v^2)-4x(u+v)=0 \Rightarrow (u^2-v^2)=x(u+v)$ so for x>0, and v≠-u≠0, we get $x=u-v$
Using this, we find
$y^2=4v(v+x)=4v(v+(u-v))=4uv\Rightarrow y=2\sqrt{uv}$ (we can omit the plus or minus, because we are considering y>0.

Staff Emeritus
Homework Helper

## Homework Statement

The question is attached in the picture.

## The Attempt at a Solution

From the question I deduced that there are 3 boundaries: red, green and blue.

Blue Boundary

This one is the easiest:

That part of the curve of y2 = 4a(a-x), which corresponds to u = a.

Red Boundary

This corresponds to x = 0, y > 0.
When x = 0,

y2 = 4u2 = 4v2, which corresponds to u = v.

Blue Boundary

This happens when y = 0, x > 0.

When y = 0,

u = 0 or u = x. ------ (1)

v = 0 or v = -x.

Not sure what is going on here...
You need to recognize that u and v are both non-negative. Then the only way to satisfy ##y^2=4v(v+x)=0## when x>0 is to have v=0.

Last edited:
unscientific
You need to recognize that u and v are both non-negative. Then the only way to satisfy ##y^2=4v(v+x)=0## when x>0 is to have v=0.

That is true, but they mentioned nothing about value of u=?

unscientific
I have a better idea..
Instead of going through and trying to formulate each of these piece by piece, try expressing x=x(u,v) and y=y(u,v)
For instance
$y^{2}=4u(u-x)=4v(v+x)\Rightarrow 4(u^2-v^2)-4x(u+v)=0 \Rightarrow (u^2-v^2)=x(u+v)$ so for x>0, and v≠-u≠0, we get $x=u-v$
Using this, we find
$y^2=4v(v+x)=4v(v+(u-v))=4uv\Rightarrow y=2\sqrt{uv}$ (we can omit the plus or minus, because we are considering y>0.

I understand that this is also a way to solve it. But sometimes the functions aren't as straightforward as these, so I would like to know the analytical method suggested here..

Staff Emeritus