# Rotating an Ellipse Anticlockwise

The quadratic equation $$5x^2-6xy+5y^2+20 \sqrt{2}x -28 \sqrt{2}y + 72 = 0$$ becomes $$(5−3 \sin(2α))x'^2 + (5+3 \sin(2α)) y'^2 - 6\cos(2α)x'y' + 4\sqrt{2}(5\cos(α)-7\sin(α))x' - 4\sqrt{2}(5\sin(α)+7\cos(α))y'+72=0$$

after applying the rotation $$x=x'cos\alpha - y'\sin\alpha, y=x'\sin\alpha + y'\cos\alpha$$.

Determine the type of the conic with equation above. Hence, determine, with respect to the $$x',y'$$ -coordinate system, the coordinates of the center, focus (foci), vertex (vertices) and where applicable asymptotes. Hence determine the coordinates of the focus (foci) in the $$x,y$$ coordinate system.

Working: The $$x'y'$$ coefficient must be $$0$$, so $$\alpha = \frac{\pi}{4}$$. This gives after simplification $$\frac{(x'-2)^2}{4} + \frac{(y'-3)^2}{1}=1$$.

Type: Ellipse

Center: $$(2,3)$$

Foci: $$(2 + \sqrt{3},3) , (2-\sqrt{3},3)$$

Vertices: $$(4,3) , (0,3)$$

No asymptotes

In $$x-y$$ coordinate system, the foci are $$(2+\sqrt{3},3),(2-\sqrt{3},3)$$.

Are these correct. I am confused babout the meaning of the finding the things like foci and center in the $$x',y'$$ coordinate system. Is that basically the $$x-y$$ axis rotated $$\frac{\pi}{4}$$ radians anticlockwise?

Type: Ellipse

Center: $$(2,3)$$

Foci: $$(2 + \sqrt{3},3) , (2-\sqrt{3},3)$$

Vertices: $$(4,3) , (0,3)$$

No asymptotes

These are correct.

In $$x-y$$ coordinate system, the foci are $$(2+\sqrt{3},3),(2-\sqrt{3},3)$$.

This is not correct. The foci are given by $$x=(2\pm\sqrt 3)\cos\frac{\pi}{4}-3\sin\frac{\pi}{4}=\frac{-\sqrt 2\pm\sqrt 6}{2}$$ $$y=(2\pm\sqrt 3)\sin\frac{\pi}{4}+3\cos\frac{\pi}{4}=\frac{5\sqrt 2\pm\sqrt 6}{2}$$

What you did is as follows :

You rotated the ellipse $$5x^2-6xy+5y^2+20 \sqrt{2}x -28 \sqrt{2}y + 72 = 0$$ by $$(-\frac{\pi}{4})$$ anticlockwise (i.e., $$\frac{\pi}{4}$$ clockwise) around the origin, and got the ellipse $$\frac{(x-2)^2}{4} + \frac{(y-3)^2}{1}=1$$. See here.

In general, we can say the followings :

• If we rotate a point $$(x,y)$$ by $$\color{red}{\theta}$$ anticlockwise around the origin, then we get $$(x\cos\theta-y\sin\theta,x\sin\theta+y\cos\theta)$$.

• If we rotate a curve $$f(x,y)=0$$ by $$\color{red}{(-\theta)}$$ anticlockwise (i.e., $$\theta$$ clockwise) around the origin, then we get a curve whose equation is $$f(x\cos\theta-y\sin\theta,x\sin\theta+y\cos\theta)=0$$.

You correctly got $$\frac{(x'-2)^2}{4} + \frac{(y'-3)^2}{1}=1$$.

In the $$x',y'$$ -coordinate system, we have, as you wrote

• Center: $$(2,3)$$

• Foci: $$(2 + \sqrt{3},3) , (2-\sqrt{3},3)$$

• Vertices: $$(4,3) , (0,3)$$

Then, you are asked to determine the coordinates of the foci in the $$x,y$$ coordinate system.

This means that you are asked to determine the coordinates of the foci of the ellipse $$5x^2-6xy+5y^2+20 \sqrt{2}x -28 \sqrt{2}y + 72 = 0\tag1$$

As I wrote, the foci of $$(1)$$ are given by $$x=(2\pm\sqrt 3)\cos\frac{\pi}{4}-3\sin\frac{\pi}{4}=\frac{-\sqrt 2\pm\sqrt 6}{2}$$ $$y=(2\pm\sqrt 3)\sin\frac{\pi}{4}+3\cos\frac{\pi}{4}=\frac{5\sqrt 2\pm\sqrt 6}{2}$$

This is because if we rotate a point $$(x,y)$$ by $$\color{red}{\theta}$$ anticlockwise around the origin, then we get $$(x\cos\theta-y\sin\theta,x\sin\theta+y\cos\theta)$$.

In the similar way, we can see that the center and the vertices of the ellipse $$(1)$$ are as follows :

• Center : $$(-\frac{\sqrt 2}{2},\frac{5\sqrt 2}{2})$$

• Vertices : $$(\frac{\sqrt 2}{2}, \frac{7\sqrt 2}{2}),(-\frac{3\sqrt 2}{2},\frac{3\sqrt 2}{2})$$

You might want to see here and here.

• So are the center and vertices in both coordinate systems the same ?
– user1071088
Mar 27 at 9:01
• I am not sure whether the foci in the $x-y$ coordinate system are correct? E.g., the point $(2,2)$ does not equal a distance $2a$ from the two foci added together. I am unsure why the coordinates of the foci aren't the same in both coordinate systems like the center and vertices ?
– user1071088
Mar 27 at 9:05
• @Nikita Mazepin : I added some explanations. Mar 27 at 9:44