# Calculus/Conic sections

Conic sections are the intersections of a surface of a cone and a plane. There are three ways to intersect. The first method is to intersect the cone vertically, which the intersection will yield a hyperbola. The second method is to intersect the cone parallel to the outermost line of the cone, which will yield a parabola. The third method is to intersect the cone horizontally or slightly horizontally, which will yield an ellipse. For more information, see Conic section. If you have knowledge on this particular subject, you can help expand here.
In future chapters, you will encounter more about conic sections. As you progress into polar coordinates, parametric equations, and three-dimensional quadric surfaces, conic sections will again be a difficult subject to discuss. In this chapter, we will only talk about the basic Cartesian-coordinate conic sections.

## Standard equations
### Ellipse

Ellipses are shapes that have an interesting property. In order to find the standard equation for the ellipse, we must know what an ellipse is. Apart from the intersection of a inclined plane and the surface of a cone, there is another way to construct an ellipse.
An ellipse is a plane curve surrounding two focal points, such that for all points on the curve, the sum of the two distances to the focal points is a constant.
Alternatively, assume points $F_{1}=(c_{x1},c_{y1})$, $F_{2}=(c_{x2},c_{y2})$, and $P=(x,y)$, where $c$ is a constant. The ellipse is a set of points that satisfies:
    $P\in \{(x,y):PF_{1}+PF_{2}=C,C\in \mathbb {R} \}$
The image on the right is the graph of an ellipse. If there is a point $P=(x,y)$, then $PF_{1}+PF_{2}=C$, where $C$ is a constant.
Knowing the defining characteristic of an ellipse, we can start find the equation.

#### Derivation

To make things simple, let's set the center of the ellipse on the origin and the points $F_{1}$ and $F_{2}$ on the $x$-axis. (see image on the right)
Since $PF_{1}+PF_{2}=C$ and $F_{1}=(c,0),F_{2}=(-c,0),P=(x,y)$, using the distance formula, we can get:
   $PF_{1}={\sqrt {(x-c)^{2}+(y-0)^{2}}}={\sqrt {x^{2}-2xc+c^{2}+y^{2}}}$
   $PF_{2}={\sqrt {(x+c)^{2}+(y-0)^{2}}}={\sqrt {x^{2}+2xc+c^{2}+y^{2}}}$
   $PF_{1}+PF_{2}={\sqrt {x^{2}-2xc+c^{2}+y^{2}}}+{\sqrt {x^{2}+2xc+c^{2}+y^{2}}}$
Now for the constant $C$. Let the length of the semi-major axis be $a$ and the length of the semi-minor axis be $b$. Imagine that point $P$ is now at the vertex, so $P_{0}=(a,0)$. At this particular point,
   $P_{0}F_{1}=a-c$
   $P_{0}F_{2}=a+c$
   $\therefore C=P_{0}F_{1}+P_{0}F_{2}=2a$
Because the definition says that $PF_{1}+PF_{2}=C$ for any $P$, we can safely say that $C=2a$. Thus, we can start solving the equation.
   ${\sqrt {x^{2}-2xc+c^{2}+y^{2}}}+{\sqrt {x^{2}+2xc+c^{2}+y^{2}}}=2a$
   $\Leftrightarrow {\sqrt {x^{2}-2xc+c^{2}+y^{2}}}=2a-{\sqrt {x^{2}+2xc+c^{2}+y^{2}}}$
   $\Leftrightarrow x^{2}-2xc+c^{2}+y^{2}=4a^{2}-4a{\sqrt {x^{2}+2xc+c^{2}+y^{2}}}+x^{2}+2xc+c^{2}+y^{2}$ (Square both sides)
   $\Leftrightarrow 4a{\sqrt {x^{2}+2xc+c^{2}+y^{2}}}=4a^{2}+4xc$ (Simplify it)
   $\Leftrightarrow a{\sqrt {x^{2}+2xc+c^{2}+y^{2}}}=a^{2}+xc$
   $\Leftrightarrow a^{2}(x^{2}+2xc+c^{2}+y^{2})=a^{4}+2a^{2}xc+x^{2}c^{2}$ (Square both sides)
   $\Leftrightarrow a^{2}x^{2}+2a^{2}xc+a^{2}c^{2}+a^{2}y^{2}=a^{4}+2a^{2}xc+x^{2}c^{2}$
   $\Leftrightarrow a^{2}x^{2}-c^{2}x^{2}+a^{2}y^{2}=a^{4}-a^{2}c^{2}$ (Simplify it)
   $\Leftrightarrow x^{2}(a^{2}-c^{2})+a^{2}y^{2}=a^{2}(a^{2}-c^{2})$ (Factor it)
   Finally, the equation is
   ${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{a^{2}-c^{2}}}=1$
This should be the equation. But $a^{2}-c^{2}$ can be further simplified. To do so, imagine again that our point $P$ is on the co-vertex, so $P_{1}=(0,b)$. Thus,
   $P_{1}F_{1}={\sqrt {c^{2}+b^{2}}}$
   $P_{1}F_{2}={\sqrt {c^{2}+b^{2}}}$
   $\therefore C=P_{1}F_{1}+P_{1}F_{2}=2{\sqrt {c^{2}+b^{2}}}$
Since we have already established that $C=2a$, we can write down an equation that links $a,b,c$ together:
   $2{\sqrt {c^{2}+b^{2}}}=2a$
   $\Leftrightarrow c^{2}+b^{2}=a^{2}$
   $\Leftrightarrow a^{2}-c^{2}=b^{2}$
Now we substitute $a^{2}-c^{2}$ as $b^{2}$, we get the final result:
${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1$
$\blacksquare$
This equation is the standard form of the ellipse. It is considered standard because all key points are on the axes.

#### Terminology and properties

We already derived the equation. So, the terminology and properties will be based on that.
   ${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1$
* Focus (plural: foci): the points $F_{1},F_{2}$ which have coordinates $(c,0),(-c,0)$ respectively. The definition gives those points their function: $PF_{1}+PF_{2}=2a$.
* Semi-major axis: the axis with length $a$.
* Semi-minor axis: the axis with length $b$, $b<a$.
* Vertex (plural: vertices): the endpoint of the semi-major axis. It has the coordinate $(\pm a,0)$.
* Co-vertex: the endpoint of the semi-minor axis. It has the coordinate $(\pm b,0)$, $b<a$.
* Center: the middle point between the two foci. It has the coordinate $(0,0)$.
Note that any changes towards the equation will change the coordinates for the key values. The coordinates above are strictly based upon the standard equation of the ellipse.
In the derivation, we stumbled upon a property, which is the relationship between the constants $a,b,c$. The property is $c^{2}=a^{2}-b^{2}$. In the derivation of the equation of the hyperbola, we will encounter this property again. However, it will be slightly different because of the signs. In the ellipse, $a>c$, so the property $c^{2}=a^{2}-b^{2}$ ensures that $a^{2}>0,b^{2}>0,c^{2}>0$. In the hyperbola, as we will see, $c>a$, which means the length of the foci to the center is larger then that of the vertices to the center. In order to ensure that $a^{2}>0,b^{2}>0,c^{2}>0$ for convenience, the property will be slightly adjusted.

#### Transformations

If we want the ellipse to be more "vertical" instead of "horizontal", the equation of the ellipse needs to be changed. To be more "vertical", the foci of the ellipse should be on the $y$-axis, having coordinates $(0,\pm c)$. Using the same method for derivation, we get:
   ${\frac {x^{2}}{b^{2}}}+{\frac {y^{2}}{a^{2}}}=1$
If we want the ellipse to translate (move without rotating) in the plane, using what we've learned in Chapter 1.2, we can modify the equation into:
   ${\frac {(x-m)^{2}}{a^{2}}}+{\frac {(y-n)^{2}}{b^{2}}}=1$, where $(m,n)$ is the center of the ellipse.

### Parabola

Parabolas can be interpreted as the more general form of the quadratic function. However, they are essentially different. While quadratic function is a function which describes a relationship between a variable and another, parabola is a curve in 2D space.
Assume there is a point $F=(p_{1},p_{2})$ and a line $l:y=mx+b$ that does not go through point $F$. The parabola is a set of points $P=(x,y)$ that satisfies:
    $P\in \{(x,y):PF=Pl\}$
To make things simple for derivation, we put the point $F$ on the $y$-axis and the vertex of the parabola on the origin (see image on the right), so $F=(0,p)$ Now, imagine that point $P$ is on the vertex, so $P_{0}=(0,0)$. Because $PF=Pl$ and $PF=p$, the equation for line $l$ is $l:y=-p$.
Now, we can start deriving the standard equation for the parabola.

#### Derivation

Since we know that $F=(0,p)$, $l:y=-p$, and $P=(x,y)$, we can solve $PF=Pl$
   $PF={\sqrt {(x-0)^{2}+(y-p)^{2}}}={\sqrt {x^{2}+y^{2}-2yp+p^{2}}}$ $Pl=y+p$
So the equation $PF=Pl$ can be turned into
   ${\sqrt {x^{2}+y^{2}-2yp+p^{2}}}=y+p$
   $\Leftrightarrow x^{2}+y^{2}-2py+p^{2}=y^{2}+2py+p^{2}$ (square both sides)
   $\Leftrightarrow x^{2}=4py$
The standard equation for a parabola is
$x^{2}=4py$
$\blacksquare$
In Chapter 1.4, we talked about the various forms of quadratic functions, which if viewed as a geometric curve, is a parabola. To make the standard equation more familiar, we can adjust it to
   $y={\frac {1}{4p}}x^{2}$

#### Terminology and properties

Similar to the ellipse, we will use the standard equation to demonstrate key terms in the parabola.
   $y={\frac {1}{4p}}x^{2}$
* Focus: the point $F$ with coordinates $(0,p)$.
* Directrix: the line $l$ with the equation $y=-p$.
* Vertex, the lowest point on the parabola with coordinates $(0,0)$.
Recall that the vertex form of the quadratic function is $y=a(x-h)^{2}+k$. We can find that ${\frac {1}{4p}}=a$.

#### Transformations

If we want the curve to face horizontally (the axis of symmetry is the $x$-axis instead of the $y$-axis), we change the coordinates for $F$ to $F=(p,0)$ and the line to $l:x=-p$. After some calculations, we get:
   $x={\frac {1}{4p}}y^{2}$
If we want to translate the parabola in the plane, using what we've learnt, we get:
   $y-k={\frac {1}{4p}}(x-h)^{2}$, where the vertex is $(h,k)$.
This resembles the vertex of the quadratic function $y=a(x-h)^{2}+k$. However, it is important to realize that parabolas and quadratic functions are fundamentally different. One is a geometric curve while the other is a transformation between two variables.

### Hyperbola

The relationship between the hyperbola and the reciprocal function is similar to that between the parabola and the quadratic function.
Assume there are points $F_{1}=(c_{x1},c_{y1})$ and $F_{2}=(c_{x2},c_{y2})$. A hyperbola is a set of points $P=(x,y)$ that satisfies:
    $P\in \{(x,y):$ |$PF_{1}-PF_{2}$| $=C,C\in \mathbb {R} \$
Again, to make things easier, we put the points $F_{1},F_{2}$ on the $x$-axis and the center on the origin, so $F_{1}=(c,0),F_{2}=(-c,0)$. Also, the length between the vertex and the center is $a$. See the image on the right for more clarification.

#### Derivation

Since we know that $F_{1}=(c,0),F_{2}=(-c,0)$, we can solve for $PF_{1}{\text{ and }}PF_{2}$.
   $PF_{1}={\sqrt {(x-c)^{2}+y^{2}}}={\sqrt {x^{2}-2xc+c^{2}+y^{2}}}$
   $PF_{2}={\sqrt {(x+c)^{2}+y^{2}}}={\sqrt {x^{2}+2xc+c^{2}+y^{2}}}$
   $|PF_{1}-PF_{2}|=|{\sqrt {x^{2}-2xc+c^{2}+y^{2}}}-{\sqrt {x^{2}+2xc+c^{2}+y^{2}}}|$
Similar to the ellipse derivation case, we need to turn the constant $C$ into something in terms of either $a$ or $c$. In this case, imagine point $P$ is on the vertex, so $P_{0}=(a,0)$.
   $P_{0}F_{1}=c-a$
   $P_{0}F_{2}=a+c$
   $|P_{0}F_{1}-P_{0}F_{2}|=|(c-a)-(a+c)|=2a$
So, $C=2a$. Now, we derive.
   $|{\sqrt {x^{2}-2xc+c^{2}+y^{2}}}-{\sqrt {x^{2}+2xc+c^{2}+y^{2}}}|=2a$
   Let's assume that $PF_{2}>PF_{1}$, which means the curve is on the positive $x$-axis, and
   ${\sqrt {x^{2}+2xc+c^{2}+y^{2}}}-{\sqrt {x^{2}-2xc+c^{2}+y^{2}}}=2a$
      ${\sqrt {x^{2}+2xc+c^{2}+y^{2}}}=2a+{\sqrt {x^{2}-2xc+c^{2}+y^{2}}}$
      $\Leftrightarrow x^{2}+2xc+c^{2}+y^{2}=4a^{2}+4a{\sqrt {x^{2}-2xc+c^{2}+y^{2}}}+x^{2}-2xc+c^{2}+y^{2}$ (square both sides)
      $\Leftrightarrow a{\sqrt {x^{2}-2xc+c^{2}+y^{2}}}=-a^{2}+xc$
      $\Leftrightarrow a^{2}x^{2}-2a^{2}xc+a^{2}c^{2}+a^{2}y^{2}=a^{4}-2a^{2}xc+x^{2}c^{2}$ (square both sides)
      $\Leftrightarrow x^{2}(a^{2}-c^{2})+a^{2}y^{2}=a^{2}(a^{2}-c^{2})$
      $\Leftrightarrow {\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{a^{2}-c^{2}}}=1$
   For convenience, we substitute $c^{2}-a^{2}$ as $b^{2}$ because looking at the graph, $c>a$, and we want our constants to be positive. So
   ${\frac {x^{2}}{a^{2}}}-{\frac {y^{2}}{b^{2}}}=1$
   What about when $PF_{2}<PF_{1}$? That means the curve is on the negative $x$-axis, and
   ${\sqrt {x^{2}-2xc+c^{2}+y^{2}}}-{\sqrt {x^{2}+2xc+c^{2}+y^{2}}}=2a$
   Using the same method, we get:
      ${\sqrt {x^{2}-2xc+c^{2}+y^{2}}}=2a+{\sqrt {x^{2}+2xc+c^{2}+y^{2}}}$
      $\Leftrightarrow x^{2}-2xc+c^{2}+y^{2}=4a^{2}+4a{\sqrt {x^{2}+2xc+c^{2}+y^{2}}}+x^{2}+2xc+c^{2}+y^{2}$
      $\Leftrightarrow a{\sqrt {x^{2}+2xc+c^{2}+y^{2}}}=a^{2}+xc$
      $\Leftrightarrow a^{2}x^{2}+2a^{2}xc+a^{2}c^{2}+a^{2}y^{2}=a^{4}+2a^{2}xc+x^{2}c^{2}$
      $\Leftrightarrow x^{2}(a^{2}-c^{2})+a^{2}y^{2}=a^{2}(a^{2}-c^{2})$
      $\Leftrightarrow {\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{a^{2}-c^{2}}}=1$
   Because we let $c^{2}-a^{2}=b^{2}$, so
   ${\frac {x^{2}}{a^{2}}}-{\frac {y^{2}}{b^{2}}}=1$
All possibilities have been discussed, and we can safely say that the equation of a hyperbola is
${\frac {x^{2}}{a^{2}}}-{\frac {y^{2}}{b^{2}}}=1$
$\blacksquare$

#### Terminology and properties

${\frac {x^{2}}{a^{2}}}-{\frac {y^{2}}{b^{2}}}=1$
* Focus (plural: foci): the points $F_{1}=(c,0),F_{2}=(-c,0)$.
* Vertex (plural: vertices): the endpoint of the semi-major axis. It has coordinates $(\pm a,0)$.
* Semi-major axis: the axis that has a length of $a$.
* Asymptote: the line where the hyperbola is approaching but never intersects.
We briefly discussed the property of the hyperbola when talking ellipses. In that discussion, we said that we have to change the property slightly to make sure that all constants are positive. In this case, since $c>a$, instead of the ellipse's $c^{2}=a^{2}-b^{2}$, we have $c^{2}=a^{2}+b^{2}$.
The asymptote is a little bit difficult to calculate because we need limits to find it.
   Since ${\frac {x^{2}}{a^{2}}}-{\frac {y^{2}}{b^{2}}}=1$, the relationship between $y$ and $x$ is:
      $y=\pm {\sqrt {{\frac {b^{2}}{a^{2}}}(x^{2}-a^{2})}}=\pm {\frac {b}{a}}{\sqrt {(x^{2}-a^{2})}}$
   Thus, the slope of the line passing through a point on the hyperbola and the center is:
      $m={\frac {\Delta y}{\Delta x}}={\frac {\pm {\frac {b}{a}}{\sqrt {x^{2}-a^{2}}}}{x}}$
   The asymptote is the line where the hyperbola is approaching but never intersects. So, we can imagine that $x$ is infinitely large to a point that it intersects the asymptote (see Chapter 2.1 and 2.3 for more).
      ${\frac {\pm {\frac {b}{a}}{\sqrt {x^{2}-a^{2}}}}{x}}$
      $x$
      $=\pm {\frac {b}{a}}$
   Formally, if we want to express "where $x$ is infinitely large", we write it like this: $\lim _{x\to \infty }{\frac {\pm {\frac {b}{a}}{\sqrt {x^{2}-a^{2}}}}{x}}=\pm {\frac {b}{a}}$. We will discuss this expression in the next unit.
   The slope of the asymptote is calculated to be $\pm {\frac {b}{a}}$.
As a result, the asymptote equation for the hyperbola is
   $y=\pm {\frac {b}{a}}x$
$\blacksquare$

#### Transformations

If we want the hyperbola to face north-south instead of east-west, changing the coordinates of key points and using the same method for derivation will yield:
   ${\frac {y^{2}}{a^{2}}}-{\frac {x^{2}}{b^{2}}}=1$
If we want to translate the hyperbola in the plane, the equation will become:
   ${\frac {(x-m)^{2}}{a^{2}}}-{\frac {(y-n)^{2}}{b^{2}}}=1$, where the center is $(m,n)$ and the asymptote is $y-n=\pm {\frac {b}{a}}(x-m)$

## Axes rotation

When we start rotating curves like conic sections, it is extremely difficult to intuitively visualize the process of rotating curves: we are more used to translation than rotation. So, instead of rotating the curve back to something like ${\frac {(x-m)^{2}}{a^{2}}}-{\frac {(y-n)^{2}}{b^{2}}}=1$, we rotate the axes to make the curve look like ${\frac {(x'-m)^{2}}{a^{2}}}-{\frac {(y'-n)^{2}}{b^{2}}}=1$. Note that $x',y'$ are the rotated coordinates.
In order to do so, we need to find out the relationship between the pre-rotation axes and the post-rotation axes. In other words, suppose there is a point with coordinates $(x,y)$. After the rotation, the coordinates are $(x',y')$. Express $(x',y')$ in terms of $x,y$.
   Now, we need to establish some coordinates and some of the properties.
   * The coordinate of the point is $(x,y)$ in a $x0y$ plane
   * The distance between the point and the origin is $r$
   * The angle between the line segment connecting the point and the origin and the positive $x$ -axis is $\alpha$
   * The axes rotated $\theta$ counterclockwise to establish a new plane: $x'0y'$
   * Bullet points 3 and 4 can help us know that the angle between the line segment connecting the point and the origin and the positive $x'$ -axis is $\alpha -\theta$
   Now, we try to evaluate $(x',y')$
      $x'=r\cos(\alpha -\theta )=r\cos \alpha \cos \theta +r\sin \alpha \sin \theta$
      $y'=r\sin(\alpha -\theta )=r\sin \alpha \cos \theta -r\cos \alpha \sin \theta$
   We can also evaluate $x,y$ in terms of $r,\alpha$:
      $x=r\cos \alpha$
      $y=r\sin \alpha$
   Then substitute into the first two equations, we get:
   $x'=x\cos \theta +y\sin \theta$
   $y'=y\cos \theta -x\sin \theta$
Finally, the coordinate for the point $(x,y)$ after the rotation is $(x\cos \theta +y\sin \theta ,y\cos \theta -x\sin \theta )$.
$\blacksquare$
The inverse transformation is $x=x'\cos \theta -y'\sin \theta ,y=x'\sin \theta +y'\cos \theta$.

## General Cartesian form

    $Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0,{\text{ where }}A,B,C,D,E,F$ are constants
The best way to get an understanding of the general form is to look at the following example. It covers most of the content in this chapter and is relatively difficult.
Example: Find the foci for this conic section with the equation $7x^{2}-2xy+7y^{2}+6{\sqrt {2}}x+6{\sqrt {2}}y-18=0$
To find the foci, we need the standard form. However, this is the general form. To make things worse, this is a rotated curve, which makes it impossible for humans to factor the equation into the translated standard form.
We know that the curve is rotated because after factoring, there is a $(ax+by)^{2}$ factor. And the translated standard form does not have that particular factor. So, we should start rotating the axes so that after the rotation, the $Bxy$ part can be cancelled.
First, we should list down what we know.
   $A=7,B=-2,C=7,D=6{\sqrt {2}},E=6{\sqrt {2}},F=-18$
Then, we start rotating the axes.
   Assume that we've rotated the axes $x0y$ into $x'0y'$ by rotating the axes by $\theta$ counterclockwise. The new rotated equation should be
   $A'x'^{2}+B'x'y'+C'y'^{2}+D'x'+E'y'+F'=0$
   Since after factoring the equation, we don't any $(ax+by)^{2}$ factor, we need to make $B'x'y'=0$. To make it simpler, we just need to make $B'=0$.
   Then, we need to express $B'$ in terms of the constants $A,B,C,D,E,F$ , and $\theta$ so that we can know how much we should rotate the axes.
      $\because x=x'\cos \theta -y'\sin \theta ,y=x'\sin \theta +y'\cos \theta$
      $\therefore Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$
   After substituting, we get:
   $A(x'\cos \theta -y'\sin \theta )^{2}+B(x'\cos \theta -y'\sin \theta )(y'\cos \theta +x'\sin \theta )+C(y'\cos \theta +x'\sin \theta )^{2}+D(x'\cos \theta -y'\sin \theta )+E(y'\cos \theta +x'\sin \theta )+F=0$
   After a lot of calculations and algebraic manipulations, we get:
      $A'=A\cos ^{2}\theta +B\cos \theta \sin \theta +C\sin ^{2}\theta$
      $B'=(A+C)2\cos \theta \sin \theta +B(\cos ^{2}\theta -\sin ^{2}\theta )$
      $C'=A\sin ^{2}\theta -B\cos \theta \sin \theta +C\cos ^{2}\theta$
      $D'=D\cos \theta +E\sin \theta$
      $E'=E\cos \theta -D\sin \theta$
      $F'=F$
   We want $B'=0$.
      $B'=(A+C)2\cos \theta \sin \theta +B(\cos ^{2}\theta -\sin ^{2}\theta )=0$
         $2\cos \theta \sin \theta =\sin 2\theta ,\cos ^{2}\theta -\sin ^{2}\theta =\cos 2\theta$
         $\cot 2\theta ={\frac {A+C}{B}}$
         For convenience, we will just solve for the simplest value.
         $\theta ={\frac {\pi }{4}}$
      $A'=7\cos ^{2}{\frac {\pi }{4}}-2\cos {\frac {\pi }{4}}\sin {\frac {\pi }{4}}+7\sin ^{2}{\frac {\pi }{4}}=6$
      $B'=(7+7)2\cos {\frac {\pi }{4}}\sin {\frac {\pi }{4}}-2(\cos ^{2}{\frac {\pi }{4}}-\sin ^{2}{\frac {\pi }{4}})=0$
      $C'=7\sin ^{2}{\frac {\pi }{4}}+2\cos {\frac {\pi }{4}}\sin {\frac {\pi }{4}}+7\cos ^{2}{\frac {\pi }{4}}=8$
      $D'=6{\sqrt {2}}\cos {\frac {\pi }{4}}+6{\sqrt {2}}\sin {\frac {\pi }{4}}=12$
      $E'=6{\sqrt {2}}\cos {\frac {\pi }{4}}-6{\sqrt {2}}\sin {\frac {\pi }{4}}=0$
      $F'=-18$
   The new rotated equation for the curve is
   $6x'^{2}+8y'^{2}+12x'-18=0$
We can now factor the equation into the translated standard form.
   $6x'^{2}+8y'^{2}+12x'-18=0$
   $\Leftrightarrow 6x'^{2}+12x'+6+8y'^{2}=24$
   $\Leftrightarrow 6(x'+1)^{2}+8y'^{2}=24$
   $\Leftrightarrow {\frac {(x'+1)^{2}}{4}}+{\frac {y'^{2}}{3}}=1$
This is the standard equation for the ellipse. Recall that in an ellipse, $c^{2}=a^{2}-b^{2}$.
Finally, we can determine the rotated coordinates of the foci.
   The rotated coordinate of the center is $(-1,0)$
   Because $c^{2}=a^{2}-b^{2}$, $c=1$.
   The rotated coordinates of the foci are
   $F_{1}'=(-2,0),F_{2}'=(0,0)$
Now, we rotate the axes back to the original state.
   $x=x'\cos \theta -y'\sin \theta ,y=x'\sin \theta +y'\cos \theta$
   $F_{1}=(-{\sqrt {2}},-{\sqrt {2}})$
   $F_{2}=(0,0)$
Therefore, the foci are $F_{1}=(-{\sqrt {2}},-{\sqrt {2}})$ and $F_{2}=(0,0)$.
$\blacksquare$