Conic section equation

Homework Statement

I have : (x)^2+(2xy)+y^2+2(2)^(1/2)x−2(2)^(1/2)y+4=0

Now when I use the rotation formulas I get : y'=(x'^2)/2 + 1

But I don't know if this makes sense....

Homework Equations

tan2β=B/(A−C)

x=x'cos β - y'sin β

y=x'cos β + y'sin β

The Attempt at a Solution

We know that β= pi/4 because B/(A−C) tends towards infinity.

So we replace pi/4 in
x=x'cos β - y'sin β

y=x'cos β + y'sin β

and put those in the long equation and simplify. Which gives us y'=((x'2)/2)+1

Does this make sense my answer

Last edited:

Related Precalculus Mathematics Homework Help News on Phys.org
Mark44
Mentor

Homework Statement

I have : (x)^2+(2xy)+y^2+2(2)^(1/2)x−2(2)^(1/2)y+4=0

Now when I use the rotation formulas I get : y'=((x'2)/2)+1

But I don't know if this makes sense....
Right. What is x'2 supposed to mean?
astrololo said:

Homework Equations

tan2β=B/(A−C)

x=x'cos β - y'sin β

y=x'cos β + y'sin β

The Attempt at a Solution

We know that β= pi/4 because B/(A−C) tends towards infinity.

So we replace pi/4 in
x=x'cos β - y'sin β

y=x'cos β + y'sin β

and put those in the long equation and simplify. Which gives us y'=((x'2)/2)+1

Does this make sense my answer

Right. What is x'2 supposed to mean?
Oh sorry, what I meant is (x'^2)/2 + 1

Mark44
Mentor
I have : (x)^2+(2xy)+y^2+2(2)^(1/2)x−2(2)^(1/2)y+4=0
This is extremely hard to read. It would be much better to write it using LaTeX, which this forum supports.

$x^2 + 2xy + y^2 + 2\sqrt{2}x - 2\sqrt{2}y + 4 = 0$

What I wrote was $x^2 + 2xy + y^2 + 2\sqrt{2}x - 2\sqrt{2}y + 4 = 0$
We have a primer on how to use LaTeX (under INFO on the menu bar) -- https://www.physicsforums.com/help/latexhelp/
astrololo said:
Now when I use the rotation formulas I get : y'=((x'2)/2)+1
Note: Corrected by OP in a later post to $y' = (1/2)(x')^2 + 1$
I don't think this is right. Your conic section is a circle, not a parabola.

Last edited:
This is extremely hard to read. It would be much better to write it using LaTeX, which this forum supports.

$x^2 + 2xy + y^2 + 2\sqrt{2}x - 2\sqrt{2}y + 4 = 0$

What I wrote was $x^2 + 2xy + y^2 + 2\sqrt{2}x - 2\sqrt{2}y + 4 = 0$
We have a primer on how to use LaTeX (under INFO on the menu bar) -- https://www.physicsforums.com/help/latexhelp/

Note: Corrected by OP in a later post to $y' = (1/2)(x')^2 + 1$
I don't think this is right. Your conic section is a circle, not a parabola.
According to wolfram Im supposed to have a parabola... I dont know what went wrong tbh

Mark44
Mentor
According to wolfram Im supposed to have a parabola... I dont know what went wrong tbh
I don't know what you entered into WA, so I can't say why WA is telling you that your equation represents a parabola. The presence of the $x^2$ and $y^2$ terms, each with a coefficient of 1, says that this is a circle. The xy term indicates that there is a rotation. You can't rotate a circle to become a parabola.

Your work on finding the angle of rotation looks OK to me. Show what you did for this part:
So we replace pi/4 in
x=x'cos β - y'sin β

y=x'cos β + y'sin β

and put those in the long equation and simplify.

I don't know what you entered into WA, so I can't say why WA is telling you that your equation represents a parabola. The presence of the $x^2$ and $y^2$ terms, each with a coefficient of 1, says that this is a circle. The xy term indicates that there is a rotation. You can't rotate a circle to become a parabola.

Your work on finding the angle of rotation looks OK to me. Show what you did for this part:
Sorry for my late reply : http://www.wolframalpha.com/input/?i=x^2+2xy+y^2+2(2)^(1/2)x-2(2)^(1/2)y+4=0

Also there is now way this can be a circle because my exercice makes it clear that its an ellipse, hyperbola or parabola

Mark44
Mentor
OK - guess I was mistaken...
MARK this problem is going to make me insane. I'm pretty sure I applied everything correctly.

Mark44
Mentor
The equation you wrote, in the transformed coordinates, is a parabola, which agrees with what WA is showing. Is there still something you don't understand?

The equation you wrote, in the transformed coordinates, is a parabola, which agrees with what WA is showing. Is there still something you don't understand?
Ok, if what I have is correct, why does wolfram show it differently in this case : http://www.wolframalpha.com/input/?i=+y=((x^2)/2)+1

This looks different from the other parabola..

Edit: Oh right I rotate it by pi/4, so I guess that I must do this myself right ? I just have no IDEA on how to rotate by graph.

Mark44
Mentor
Ok, if what I have is correct, why does wolfram show it differently in this case : http://www.wolframalpha.com/input/?i=+y=((x^2)/2)+1

This looks different from the other parabola..

Edit: Oh right I rotate it by pi/4, so I guess that I must do this myself right ? I just have no IDEA on how to rotate by graph.
Think about it this way: instead of rotating the graph by $\pi/4$ counterclockwise, rotate the axes $\pi/4$ clockwise.

Think about it this way: instead of rotating the graph by $\pi/4$ counterclockwise, rotate the axes $\pi/4$ clockwise.
The problem isn't the rotating, the problem is how do I graph a rotated parabola ? I have no idea on how you do this becaus Ive never done it. How do I even begin ? And btw I Must rotate it counter clock wise because my exercice offers me only the second quadrant. How do I find the vertex and the focus?

Mark44
Mentor
The centerline of the parabola lies along the line y = -x. The vertex of the parabola in (x', y') coordinates is at (0, 1). If this point is rotated by $\pi/4$ where does it go? Two other points in the (x', y') coordinates are at (1, 3/2) and (-1, 3/2). Figure out where these points rotate to. Then you have three points and the centerline in (x, y) coordinates, so you should be able to get a rough graph of the parabola.

The centerline of the parabola lies along the line y = -x. The vertex of the parabola in (x', y') coordinates is at (0, 1). If this point is rotated by $\pi/4$ where does it go? Two other points in the (x', y') coordinates are at (1, 3/2) and (-1, 3/2). Figure out where these points rotate to. Then you have three points and the centerline in (x, y) coordinates, so you should be able to get a rough graph of the parabola.
These two other point you took them randomly ? Also, I need to determine what is the focus in this problem. How do u do that

Mark44
Mentor
These two other point you took them randomly ? Also, I need to determine what is the focus in this problem. How do u do that
Not randomly. I used x' values of -1, 0, and 1, and calculated the y' values.

To find the focus, see https://en.wikipedia.org/wiki/Parabola#Focal_length. You'll need to figure out where the focus is for y' = (1/2)(x')2, and then translate it up by 1 unit. Finally, rotate this point by $\pi/4$ to get the (x, y) coordinates.

Not randomly. I used x' values of -1, 0, and 1, and calculated the y' values.

To find the focus, see https://en.wikipedia.org/wiki/Parabola#Focal_length. You'll need to figure out where the focus is for y' = (1/2)(x')2, and then translate it up by 1 unit. Finally, rotate this point by $\pi/4$ to get the (x, y) coordinates.
Ok, but one thing which I don't understand. When I find my normal point of x',y' and when I rotate these points, is there some analytical method which gives me the rotation of these points? Or I do all of this by drawing ?

Not randomly. I used x' values of -1, 0, and 1, and calculated the y' values.

To find the focus, see https://en.wikipedia.org/wiki/Parabola#Focal_length. You'll need to figure out where the focus is for y' = (1/2)(x')2, and then translate it up by 1 unit. Finally, rotate this point by $\pi/4$ to get the (x, y) coordinates.
Ok I was able to find (0,3/2) for my focus and I was able to draw my parabola. Now it's the rotation which I need to figure out how to do.

Mark44
Mentor
Ok, but one thing which I don't understand. When I find my normal point of x',y' and when I rotate these points, is there some analytical method which gives me the rotation of these points? Or I do all of this by drawing ?
You can find them by drawing them, or you can identify the points using simple trig. For example, the vertex in (x', y') coordinates is at (0, 1). If you rotate this point CCW, you get $(-\sqrt{2}/2, \sqrt{2}/2)$, or about (-.71, .71). The other points will be a little bit harder, but keep in mind that the radial distance (distance between the point and the origin) doesn't change.

You can find them by drawing them, or you can identify the points using simple trig. For example, the vertex in (x', y') coordinates is at (0, 1). If you rotate this point CCW, you get $(-\sqrt{2}/2, \sqrt{2}/2)$, or about (-.71, .71). The other points will be a little bit harder, but keep in mind that the radial distance (distance between the point and the origin) doesn't change.
I hope that I don't bother you with my one thousand questions. First, can you tell me the analytic method ? Like some wiki page. Also, the radial distance is the distance between any point of the parabola and the origin right ?

I need to show how I obtained my data so I need the analytical method.

Mark44
Mentor
I hope that I don't bother you with my one thousand questions. First, can you tell me the analytic method ? Like some wiki page.
If there is a wiki page for it, it would be under rotation transformations.
astrololo said:
Also, the radial distance is the distance between any point of the parabola and the origin right ?
Yes. In a rotation, the distance from the origin doesn't change.
astrololo said:
I need to show how I obtained my data so I need the analytical method.
You can use a matrix to rotate a vector, similar to what you did in post #1, but I don't know if you have learned that technique yet.

If there is a wiki page for it, it would be under rotation transformations.
Yes. In a rotation, the distance from the origin doesn't change.
Ok, I finally understand. We did a rotation which permitted me to identify a parabola. Now I must redo a rotation but inversly. so I found these equations : x=x'cos theta - y' sin theta

Which seems to work because I obtained the correct vertex of -1/2^(1/2), 1/2^(1/2)

Ill continue and see where it leads. Thank you

If there is a wiki page for it, it would be under rotation transformations.
Yes. In a rotation, the distance from the origin doesn't change.

You can use a matrix to rotate a vector, similar to what you did in post #1, but I don't know if you have learned that technique yet.
Hey can you try 1,3/2 for me please ? Im not sure if I did my part correctly. I obtian -(2)^(1/2)/4 although I dont think its correct

Mark44
Mentor
Hey can you try 1,3/2 for me please ? Im not sure if I did my part correctly. I obtian -(2)^(1/2)/4 although I dont think its correct
In polar coordinates, (1, 3/2) is $(\sqrt{13}/2, \theta)$ where $\theta = \arctan(1.5) \approx .98$ (radians). If you rotate this by $\pi/4$ you get $(\sqrt{13}/2, \theta + \pi/4)$, again in polar form. Can you work that out?

In polar coordinates, (1, 3/2) is $(\sqrt{13}/2, \theta)$ where $\theta = \arctan(1.5) \approx .98$ (radians). If you rotate this by $\pi/4$ you get $(\sqrt{13}/2, \theta + \pi/4)$, again in polar form. Can you work that out?
Forget it, I was able to make it work. In fact, I found an easier way. I rotate one coordinate x and I replace it in my long equation to find the y. (Using worlfram to find the y)