Solving Equation

[S_100]

Given that s = x ² + y ²and t = 2xy, find expressions for x and y in terms of s and t.


My solution:

At first I thought this was paramatetric equation.But then I was sure for this reason (please correct if wrong) : that a set of the equations of a parametric can only contain three variables, with the two being the cartesian ones and the third being called the parameter.

Q1) So, does that imply s and t are infact constants? and so can be solved like simulraneous equations where it can be solved by substitution?.
y= t/2x,

s = x ² + (t/2x) ²

Then rearrange to get a quartic/ diguised quadratic and find x using the quadratic forumula.--->
sub x into y= t/2x

Q2) Is there not quicker method of doing this?

 

[Steven G]

Not saying that this will help but have you noticed that (x+y)^2 =x^2 +y^2 + 2xy = s + t.
That would be the 1st thing that I would observe. Does it help?
Can you go to polar coordinates? Will this help.

You got to s = x² + (t/2x)² but then s if a function of x and t. Can you get rid of t?

Keep trying! You will get it.

 

[Dr.Peterson]

Given that s = x ² + y ²and t = 2xy, find expressions for x and y in terms of s and t.

My solution:

At first I thought this was paramatetric equation.But then I was sure for this reason (please correct if wrong) : that a set of the equations of a parametric can only contain three variables, with the two being the cartesian ones and the third being called the parameter.

Q1) So, does that imply s and t are infact constants? and so can be solved like simulraneous equations where it can be solved by substitution?.
y= t/2x,

s = x ² + (t/2x) ²

Then rearrange to get a quartic/ diguised quadratic and find x using the quadratic forumula.--->
sub x into y= t/2x

Q2) Is there not quicker method of doing this?

What you have here are two pairs of variables, (x,y) and (s,t). As written, you can think of x and y as parameters.

But that doesn't matter. Your task can be thought of as if x and y are constants and you need to solve for s and t in terms of those constants. So if you knew that x² + y² = 13 and 2xy = 12, how would you solve for x and y? The quadratic formula may be part of your work, or may not.

 

[HallsofIvy]

s = x ² + (t/2x) ²

Yes, you get a quadratic. That's not so awful is it? Tedious but straightforward. For the moment, let \(\displaystyle p= x^2\) so the equation becomes \(\displaystyle s= p+ \frac{t}{2p}\). multiply on both sides by p: \(\displaystyle sp= p^2+ \frac{t}{2}\) or \(\displaystyle p^2- sp+ \frac{t}{2}= 0\). By the quadratic formula, \(\displaystyle p= x^2= \frac{s\pm\sqrt{s^2- 2t}}{2}\). Take the square root of that: \(\displaystyle x= \pm\sqrt{\frac{s\pm\sqrt{s^2- 2t}}{2}}\).

It should be clear, by symmetry, that y is exactly the same function of s and t.

 

[Dr.Peterson]

Of course, it isn't a function; there are four possible values. And it's likely that the choice of signs for y will depend on the choice of signs for x.

Of course, the problem didn't ask for functions, just for expressions ...

 

[Steven G]

Yes, you get a quadratic. That's not so awful is it? Tedious but straightforward. For the moment, let \(\displaystyle p= x^2\) so the equation becomes \(\displaystyle s= p+ \frac{t}{2p}\). multiply on both sides by p: \(\displaystyle sp= p^2+ \frac{t}{2}\) or \(\displaystyle p^2- sp+ \frac{t}{2}= 0\). By the quadratic formula, \(\displaystyle p= x^2= \frac{s\pm\sqrt{s^2- 2t}}{2}\). Take the square root of that: \(\displaystyle x= \pm\sqrt{\frac{s\pm\sqrt{s^2- 2t}}{2}}\).

It should be clear, by symmetry, that y is exactly the same function of s and t.

Prof Halls, I think that you have been a little tired lately or you wanted the OP to catch your error. You have a mistake in the top line for s.

 

[hoosie]

A good way to approach this question is to think of s = x² + y² and t = 2xy as rules representing a circle and an hyperbola where s and t are real number constants.
Let s = 9 to give a circle of radius 3.
x² + y² = 9 passes through (1, 2√2) since 1² + (2√2)² = 9
For (1, 2√2) to lie on hyperbola t = 2xy
let t = 2(1)(2√2)
= 4√2
So we have equation 1: x² + y² = 9 and equation 2: 2xy = 4√2

Now let’s find x and y in terms of 9 and 4√2
From equation 2: y = 2√2/x
Substitute in equation 1: x² + (2√2/x)² = 9
x² + (2√2)²/x² = 9
Multiply both sides by x²
x⁴ + (2√2)² = 9x²
x⁴ - 9x² + (2√2)² = 0

Let a = x²
Solve: a² - 9a + (2√2)² = 0
By quadratic formula:
a = {-(-9) ± √[(-9)² - 4(1)(2√2)²]} / 2(1)
a = {9± √[9² - 4(2√2)²]} / 2
∴ x² = {9± √[9² - 4(2√2)²]} / 2
x = ± √[{9± √[9² - 4(2√2)²]} / 2]

y = 2√2/x
= 2√2 ÷ ± √[{9± √[9² - 4(2√2)²]} / 2]
= ± 2√2 √[2 / {9± √[9² - 4(2√2)²]}]

Since s = 9 and t = 4√2 it follows:
x = ± √[{s± √[s² - 4(t/2)²]} / 2]
= ± √[{s± √(s² - t²)} / 2]

y = ± t/2 √[2 / {s± √[s² - 4(t/2)²]}]
= ± t/2 √[2 / {s± √[s² - t²]}]

Note: When we substituted y = 2√2/x into x² + y² = 9 and solved for x we found:

x = ± √[{9± √[9² - 4(2√2)²]} / 2]
y = = ± 2√2 √[2 / {9± √[9² - 4(2√2)²]}]

These answers show that the circle x² + y² = 9 intersects the hyperbola y = 2√2/x at four points.
See if you can simplify these expressions for x and y and find the coordinates of the four points.
Graph on the same set of axes:
y = √(9 - x²) top half of circle
y = -√(9 - x²) bottom half of circle
y = 2√2/x hyperbola with two curves
Show the four intersecting points on your graph.