11.6 Hooked on Conics Again

Rotation of Axes

Consider the $x$- and $y$-axes below along with the dashed $x^{\prime }$- and $y^{\prime }$-axes obtained by rotating the $x$- and $y$-axes counter-clockwise through an angle $\theta$ and consider the point $P(x,y)$. The coordinates $(x,y)$ are rectangular coordinates and are based on the $x$- and $y$-axes. Suppose we wished to find rectangular coordinates based on the $x^{\prime }$- and $y^{\prime }$-axes. That is, we wish to determine $P(x^{\prime },y^{\prime })$. While this seems like a formidable challenge, it is nearly trivial if we use polar coordinates. Consider the angle $\varphi$ whose initial side is the positive $x^{\prime }$-axis and whose terminal side contains the point $P$.

We relate $P(x,y)$ and $P(x^{\prime },y^{\prime })$ by converting them to polar coordinates. Converting $P(x,y)$ to polar coordinates with $r>0$ yields $x=r\cos (\theta +\varphi )$ and $y=r\sin (\theta +\varphi )$. To convert the point $P(x^{\prime },y^{\prime })$ into polar coordinates, we first match the polar axis with the positive $x^{\prime }$-axis, choose the same $r>0$ (since the origin is the same in both systems) and get $x^{\prime }=r\cos (\varphi )$ and $y^{\prime }=r\sin (\varphi )$. Using the sum formulas for sine and cosine, we have

$$\begin{array}{cccc}\hfill x& =& r\cos (\theta +\varphi )\hfill & \\ & =& r\cos (\theta )\cos (\varphi )-r\sin (\theta )\sin (\varphi )\hfill & \text{Sum formula for cosine}\hfill \\ & =& (r\cos (\varphi ))\cos (\theta )-(r\sin (\varphi ))\sin (\theta )\hfill & \\ & =& x^{\prime }\cos (\theta )-y^{\prime }\sin (\theta )\hfill & \text{Since }{x}^{\prime }=r\cos (\varphi )\text{ and }{y}^{\prime }=r\sin (\varphi )\hfill \end{array}$$

Similarly, using the sum formula for sine we get $y=x^{\prime }\sin (\theta )+y^{\prime }\cos (\theta )$. These equations enable us to easily convert points with $x^{\prime }{y}^{\prime }$-coordinates back into $xy$-coordinates. They also enable us to easily convert equations in the variables $x$ and $y$ into equations in the variables in terms of $x^{\prime }$ and $y^{\prime }$.¹ If we want equations which enable us to convert points with $xy$-coordinates into $x^{\prime }{y}^{\prime }$-coordinates, we need to solve the system

$$\{\begin{array}{ccc}\hfill x^{\prime }\cos (\theta )-y^{\prime }\sin (\theta )& =& x\hfill \\ \hfill x^{\prime }\sin (\theta )+y^{\prime }\cos (\theta )& =& y\hfill \end{array}$$

for $x^{\prime }$ and $y^{\prime }$. Perhaps the cleanest way² to solve this system is to write it as a matrix equation. Using the machinery developed in Section, we write the above system as the matrix equation $A{X}^{\prime }=X$ where

$$\begin{array}{ccc}A=\left[\begin{array}{cc}\hfill \cos (\theta )& \hfill -\sin (\theta )\\ \hfill \sin (\theta )& \hfill \cos (\theta )\end{array}\right],& X^{\prime }=\left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right],& X=\left[\begin{array}{c}x\\ y\end{array}\right]\end{array}$$

Since $$det(A)=(\cos (\theta ))(\cos (\theta ))-(-\sin (\theta ))(\sin (\theta ))={\cos }^2(\theta )+{\sin }^2(\theta )=1$$, the determinant of $A$ is not zero so $A$ is invertible and $X^{\prime }=A^{-1}X$. Using the formula given in Equation with $det(A)=1$, we find

$$A^{-1}=\left[\begin{array}{cc}\hfill \cos (\theta )& \hfill \sin (\theta )\\ \hfill -\sin (\theta )& \hfill \cos (\theta )\end{array}\right]$$

so that

$$\begin{array}{ccc}\hfill X^{\prime }& =& A^{-1}X\hfill \\ \hfill \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]& =& \left[\begin{array}{cc}\hfill \cos (\theta )& \hfill \sin (\theta )\\ \hfill -\sin (\theta )& \hfill \cos (\theta )\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]\hfill \\ \hfill \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]& =& \left[\begin{array}{c}x\cos (\theta )+y\sin (\theta )\\ -x\sin (\theta )+y\cos (\theta )\end{array}\right]\hfill \end{array}$$

From which we get $x^{\prime }=x\cos (\theta )+y\sin (\theta )$ and $y^{\prime }=-x\sin (\theta )+y\cos (\theta )$. To summarize,

We put the formulas in Theorem to good use in the following example.

The elimination of the troublesome `$xy$' term from the equation $21x^2+10xy\sqrt{3}+31y^2=144$ in Example Example 1 number allowed us to graph the equation by hand using what we learned in Chapter. It is natural to wonder if we can always do this. That is, given an equation of the form $$A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$$, with $B\ne 0$, is there an angle $\theta$ so that if we rotate the $x$ and $y$-axes counter-clockwise through that angle $\theta$, the equation in the rotated variables $x^{\prime }$ and $y^{\prime }$ contains no $x^{\prime }{y}^{\prime }$ term? To explore this conjecture, we make the usual substitutions $x=x^{\prime }\cos (\theta )-y^{\prime }\sin (\theta )$ and $y=x^{\prime }\sin (\theta )+y^{\prime }\cos (\theta )$ into the equation $$A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$$ and set the coefficient of the $x^{\prime }{y}^{\prime }$ term equal to $0$. Terms containing $x^{\prime }{y}^{\prime }$ in this expression will come from the first three terms of the equation: $A{x}^2$, $Bxy$ and $C{y}^2$. We leave it to the reader to verify that

$$\begin{array}{ccc}\hfill x^2& =& {(x^{\prime })}^2{\cos }^2(\theta )-2x^{\prime }{y}^{\prime }\cos (\theta )\sin (\theta )+{(y^{\prime })}^2\sin (\theta )\hfill \\ \hfill xy& =& {(x^{\prime })}^2\cos (\theta )\sin (\theta )+x^{\prime }{y}^{\prime }\left({\cos }^2(\theta )-{\sin }^2(\theta )\right)-{(y^{\prime })}^2\cos (\theta )\sin (\theta )\hfill \\ \hfill y^2& =& {(x^{\prime })}^2{\sin }^2(\theta )+2x^{\prime }{y}^{\prime }\cos (\theta )\sin (\theta )+{(y^{\prime })}^2{\cos }^2(\theta )\hfill \end{array}$$

The contribution to the $x^{\prime }{y}^{\prime }$-term from $A{x}^2$ is $-2A\cos (\theta )\sin (\theta )$, from $Bxy$ it is $B\left({\cos }^2(\theta )-{\sin }^2(\theta )\right)$, and from $C{y}^2$ it is $2C\cos (\theta )\sin (\theta )$. Equating the $x^{\prime }{y}^{\prime }$-term to $0$, we get

$$\begin{array}{cccc}\hfill -2A\cos (\theta )\sin (\theta )+B\left({\cos }^2(\theta )-{\sin }^2(\theta )\right)+2C\cos (\theta )\sin (\theta )& =& 0\hfill & \\ \hfill -A\sin (2\theta )+B\cos (2\theta )+C\sin (2\theta )& =& 0\hfill & \text{Double Angle Identities}\hfill \end{array}$$

From this, we get $$B\cos (2\theta )=(A-C)\sin (2\theta )$$, and our goal is to solve for $\theta$ in terms of the coefficients $A$, $B$ and $C$. Since we are assuming $B\ne 0$, we can divide both sides of this equation by $B$. To solve for $\theta$ we would like to divide both sides of the equation by $\sin (2\theta )$, provided of course that we have assurances that $\sin (2\theta )\ne 0$. If $\sin (2\theta )=0$, then we would have $B\cos (2\theta )=0$, and since $B\ne 0$, this would force $\cos (2\theta )=0$. Since no angle $\theta$ can have both $\sin (2\theta )=0$ and $\cos (2\theta )=0$, we can safely assume³ $\sin (2\theta )\ne 0$. We get $\frac{\cos (2\theta )}{\sin (2\theta )}=\frac{A-C}{B}$, or $\cot (2\theta )=\frac{A-C}{B}$. We have just proved the following theorem.

We put Theorem to good use in the following example.

We note that even though the coefficients of $x^2$ and $y^2$ were both positive numbers in parts and of Example Example 2, the graph in part turned out to be a hyperbola and the graph in part worked out to be a parabola. Whereas in Chapter, we could easily pick out which conic section we were dealing with based on the presence (or absence) of quadratic terms and their coefficients, Example Example 2 demonstrates that all bets are off when it comes to conics with an $xy$ term which require rotation of axes to put them into a more standard form. Nevertheless, it is possible to determine which conic section we have by looking at a special, familiar combination of the coefficients of the quadratic terms. We have the following theorem.

As you may expect, the quantity $B^2-4AC$ mentioned in Theorem is called the discriminant of the conic section. While we will not attempt to explain the deep Mathematics which produces this `coincidence', we will at least work through the proof of Theorem mechanically to show that it is true.⁶ First note that if the coefficient $B=0$ in the equation $$A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$$, Theorem reduces to the result presented in Exercise in Section, so we proceed here under the assumption that $B\ne 0$. We rotate the $xy$-axes counter-clockwise through an angle $\theta$ which satisfies $\cot (2\theta )=\frac{A-C}{B}$ to produce an equation with no $x^{\prime }{y}^{\prime }$-term in accordance with Theorem: $$A^{\prime }{(x^{\prime })}^2+C{(y^{\prime })}^2+D{x}^{\prime }+E{y}^{\prime }+F^{\prime }=0$$. In this form, we can invoke Exercise in Section once more using the product $A^{\prime }{C}^{\prime }$. Our goal is to find the product $A^{\prime }{C}^{\prime }$ in terms of the coefficients $A$, $B$ and $C$ in the original equation. To that end, we make the usual substitutions $x=x^{\prime }\cos (\theta )-y^{\prime }\sin (\theta )$ $y=x^{\prime }\sin (\theta )+y^{\prime }\cos (\theta )$ into $$A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$$. We leave it to the reader to show that, after gathering like terms, the coefficient $A^{\prime }$ on ${(x^{\prime })}^2$ and the coefficient $C^{\prime }$ on ${(y^{\prime })}^2$ are

$$\begin{array}{cccc}\hfill A^{\prime }& =& A{\cos }^2(\theta )+B\cos (\theta )\sin (\theta )+C{\sin }^2(\theta )\hfill & \\ \hfill C^{\prime }& =& A{\sin }^2(\theta )-B\cos (\theta )\sin (\theta )+C{\cos }^2(\theta )\hfill & \end{array}$$

In order to make use of the condition $\cot (2\theta )=\frac{A-C}{B}$, we rewrite our formulas for $A^{\prime }$ and $C^{\prime }$ using the power reduction formulas. After some regrouping, we get

$$\begin{array}{cccc}\hfill 2A^{\prime }& =& \left[(A+C)+(A-C)\cos (2\theta )\right]+B\sin (2\theta )\hfill & \\ \hfill 2C^{\prime }& =& \left[(A+C)-(A-C)\cos (2\theta )\right]-B\sin (2\theta )\hfill & \end{array}$$

Next, we try to make sense of the product

$$(2A^{\prime })(2C^{\prime })=\left\{\left[(A+C)+(A-C)\cos (2\theta )\right]+B\sin (2\theta )\right\}\left\{\left[(A+C)-(A-C)\cos (2\theta )\right]-B\sin (2\theta )\right\}$$

We break this product into pieces. First, we use the difference of squares to multiply the `first' quantities in each factor to get

$$\begin{array}{ccc}\hfill \left[(A+C)+(A-C)\cos (2\theta )\right]\left[(A+C)-(A-C)\cos (2\theta )\right]& =& {(A+C)}^2-{(A-C)}^2{\cos }^2(2\theta )\hfill \end{array}$$

Next, we add the product of the `outer' and `inner' quantities in each factor to get

$$\begin{array}{ccc}\hfill -B\sin (2\theta )\left[(A+C)+(A-C)\cos (2\theta )\right]& & \\ \hfill +B\sin (2\theta )\left[(A+C)-(A-C)\cos (2\theta )\right]& =& -2B(A-C)\cos (2\theta )\sin (2\theta )\hfill \end{array}$$

The product of the `last' quantity in each factor is $$(B\sin (2\theta ))(-B\sin (2\theta ))=-B^2{\sin }^2(2\theta )$$. Putting all of this together yields

$$\begin{array}{ccc}\hfill 4A^{\prime }{C}^{\prime }& =& {(A+C)}^2-{(A-C)}^2{\cos }^2(2\theta )-2B(A-C)\cos (2\theta )\sin (2\theta )-B^2{\sin }^2(2\theta )\hfill \end{array}$$

From $\cot (2\theta )=\frac{A-C}{B}$, we get $\frac{\cos (2\theta )}{\sin (2\theta )}=\frac{A-C}{B}$, or $$(A-C)\sin (2\theta )=B\cos (2\theta )$$. We use this substitution twice along with the Pythagorean Identity ${\cos }^2(2\theta )=1-{\sin }^2(2\theta )$ to get

$$\begin{array}{ccc}\hfill 4A^{\prime }{C}^{\prime }& =& {(A+C)}^2-{(A-C)}^2{\cos }^2(2\theta )-2B(A-C)\cos (2\theta )\sin (2\theta )-B^2{\sin }^2(2\theta )\hfill \\ & =& {(A+C)}^2-{(A-C)}^2\left[1-{\sin }^2(2\theta )\right]-2B\cos (2\theta )B\cos (2\theta )-B^2{\sin }^2(2\theta )\hfill \\ & =& {(A+C)}^2-{(A-C)}^2+{(A-C)}^2{\sin }^2(2\theta )-2B^2{\cos }^2(2\theta )-B^2{\sin }^2(2\theta )\hfill \\ & =& {(A+C)}^2-{(A-C)}^2+{\left[(A-C)\sin (2\theta )\right]}^2-2B^2{\cos }^2(2\theta )-B^2{\sin }^2(2\theta )\hfill \\ & =& {(A+C)}^2-{(A-C)}^2+{\left[B\cos (2\theta )\right]}^2-2B^2{\cos }^2(2\theta )-B^2{\sin }^2(2\theta )\hfill \\ & =& {(A+C)}^2-{(A-C)}^2+B^2{\cos }^2(2\theta )-2B^2{\cos }^2(2\theta )-B^2{\sin }^2(2\theta )\hfill \\ & =& {(A+C)}^2-{(A-C)}^2-B^2{\cos }^2(2\theta )-B^2{\sin }^2(2\theta )\hfill \\ & =& {(A+C)}^2-{(A-C)}^2-B^2\left[{\cos }^2(2\theta )+{\sin }^2(2\theta )\right]\hfill \\ & =& {(A+C)}^2-{(A-C)}^2-B^2\hfill \\ & =& \left(A^2+2AC+C^2\right)-\left(A^2-2AC+C^2\right)-B^2\hfill \\ & =& 4AC-B^2\hfill \end{array}$$

Hence, $B^2-4AC=-4A^{\prime }{C}^{\prime }$, so the quantity $B^2-4AC$ has the opposite sign of $A^{\prime }{C}^{\prime }$. The result now follows by applying Exercise in Section.

The Polar Form of Conics

In this subsection, we start from scratch to reintroduce the conic sections from a more unified perspective. We have our `new' definition below.

We have seen the notions of focus and directrix before in the definition of a parabola, Definition. There, a parabola is defined as the set of points equidistant from the focus and directrix, giving an eccentricity $e=1$ according to Definition. We have also seen the concept of eccentricity before. It was introduced for ellipses in Definition in Section, and later extended to hyperbolas in Exercise in Section. There, $e$ was also defined as a ratio of distances, though in these cases the distances involved were measurements from the center to a focus and from the center to a vertex. One way to reconcile the `old' ideas of focus, directrix and eccentricity with the `new' ones presented in Definition is to derive equations for the conic sections using Definition and compare these parameters with what we know from Chapter. We begin by assuming the conic section has eccentricity $e$, a focus $F$ at the origin and that the directrix is the vertical line $x=-d$ as in the figure below.

Using a polar coordinate representation $P(r,\theta )$ for a point on the conic with $r>0$, we get

$$e=\frac{\text{the distance from }P\text{ to }F}{\text{the distance from }P\text{ to }L}=\frac{r}{d+r\cos (\theta )}$$

so that $r=e(d+r\cos (\theta ))$. Solving this equation for $r$, yields

$$r=\frac{ed}{1-e\cos (\theta )}$$

At this point, we convert the equation $r=e(d+r\cos (\theta ))$ back into a rectangular equation in the variables $x$ and $y$. If $e>0$, but $e\ne 1$, the usual conversion process outlined in Section gives⁷

$$\left(\frac{{\left(1-e^2\right)}^2}{e^2{d}^2}\right){\left(x-\frac{e^{2}d}{1-e^2}\right)}^2+\left(\frac{1-e^2}{e^2{d}^2}\right)y^2=1$$

We leave it to the reader to show if $0<e<1$, this is the equation of an ellipse centered at $(\frac{e^{2}d}{1-e^2},0)$ with major axis along the $x$-axis. Using the notation from Section, we have $a^2=\frac{e^2{d}^2}{{\left(1-e^2\right)}^2}$ and $b^2=\frac{e^2{d}^2}{1-e^2}$, so the major axis has length $\frac{2ed}{1-e^2}$ and the minor axis has length $\frac{2ed}{\sqrt{1-e^2}}$. Moreover, we find that one focus is $(0,0)$ and working through the formula given in Definition gives the eccentricity to be $e$, as required. If $e>1$, then the equation generates a hyperbola with center $(\frac{e^{2}d}{1-e^2},0)$ whose transverse axis lies along the $x$-axis. Since such hyperbolas have the form $\frac{{(x-h)}^2}{a^2}-\frac{y^2}{b^2}=1$, we need to take the opposite reciprocal of the coefficient of $y^2$ to find $b^2$. We get⁸ $a^2=\frac{e^2{d}^2}{{\left(1-e^2\right)}^2}=\frac{e^2{d}^2}{{\left(e^2-1\right)}^2}$ and $b^2=-\frac{e^2{d}^2}{1-e^2}=\frac{e^2{d}^2}{e^2-1}$, so the transverse axis has length $\frac{2ed}{e^2-1}$ and the conjugate axis has length $\frac{2ed}{\sqrt{e^2-1}}$. Additionally, we verify that one focus is at $(0,0)$, and the formula given in Exercise in Section gives the eccentricity is $e$ in this case as well. If $e=1$, the equation $r=\frac{ed}{1-e\cos (\theta )}$ reduces to $r=\frac{d}{1-\cos (\theta )}$ which gives the rectangular equation $y^2=2d\left(x+\frac{d}{2}\right)$. This is a parabola with vertex $(-\frac{d}{2},0)$ opening to the right. In the language of Section, $4p=2d$ so $p=\frac{d}{2}$, the focus is $(0,0)$, the focal diameter is $2d$ and the directrix is $x=-d$, as required. Hence, we have shown that in all cases, our `new' understanding of `conic section', `focus', `eccentricity' and `directrix' as presented in Definition correspond with the `old' definitions given in Chapter.

Before we summarize our findings, we note that in order to arrive at our general equation of a conic $r=\frac{ed}{1-e\cos (\theta )}$, we assumed that the directrix was the line $x=-d$ for $d>0$. We could have just as easily chosen the directrix to be $x=d$, $y=-d$ or $y=d$. As the reader can verify, in these cases we obtain the forms $r=\frac{ed}{1+e\cos (\theta )}$, $r=\frac{ed}{1-e\sin (\theta )}$ and $r=\frac{ed}{1+e\sin (\theta )}$, respectively. The key thing to remember is that in any of these cases, the directrix is always perpendicular to the major axis of an ellipse and it is always perpendicular to the transverse axis of the hyperbola. For parabolas, knowing the focus is $(0,0)$ and the directrix also tells us which way the parabola opens. We have established the following theorem.

We test out Theorem in the next example.

In light of Section, the reader may wonder what the rotated form of the conic sections would look like in polar form. We know from Exercise in Section that replacing $\theta$ with $(\theta -\varphi )$ in an expression $r=f(\theta )$ rotates the graph of $r=f(\theta )$ counter-clockwise by an angle $\varphi$. For instance, to graph $r=\frac{4}{1-\sin \left(\theta -\frac{\pi }{4}\right)}$ all we need to do is rotate the graph of $r=\frac{4}{1-\sin \left(\theta \right)}$, which we obtained in Example Example 4 number, counter-clockwise by $\frac{\pi }{4}$ radians, as shown below.

Using rotations, we can greatly simplify the form of the conic sections presented in Theorem, since any three of the forms given there can be obtained from the fourth by rotating through some multiple of $\frac{\pi }{2}$. Since rotations do not affect lengths, all of the formulas for lengths Theorem remain intact. In the theorem below, we also generalize our formula for conic sections to include circles centered at the origin by extending the concept of eccentricity to include $e=0$. We conclude this section with the statement of the following theorem.

Exercises

Graph the following equations.

1. $$x^2+2xy+y^2-x\sqrt{2}+y\sqrt{2}-6=0$$
2. $$7x^2-4xy\sqrt{3}+3y^2-2x-2y\sqrt{3}-5=0$$
3. $$5x^2+6xy+5y^2-4\sqrt{2}x+4\sqrt{2}y=0$$
4. $$x^2+2\sqrt{3}xy+3y^2+2\sqrt{3}x-2y-16=0$$
5. $$13x^2-34xy\sqrt{3}+47y^2-64=0$$
6. $x^2-2\sqrt{3}xy-y^2+8=0$
7. $$x^2-4xy+4y^2-2x\sqrt{5}-y\sqrt{5}=0$$
8. $$8x^2+12xy+17y^2-20=0$$
9. $r=\frac{2}{1-\cos (\theta )}$
10. $r=\frac{3}{2+\sin (\theta )}$
11. $r=\frac{3}{2-\cos (\theta )}$
12. $r=\frac{2}{1+\sin (\theta )}$
13. $r=\frac{4}{1+3\cos (\theta )}$
14. $r=\frac{2}{1-2\sin (\theta )}$
15. $r=\frac{2}{1+\sin (\theta -\frac{\pi }{3})}$
16. $r=\frac{6}{3-\cos \left(\theta +\frac{\pi }{4}\right)}$
17. Show the matrix from Example in Section is none other than $A\left(\frac{\pi }{4}\right)$.
18. Discuss with your classmates how to use $A(\theta )$ to rotate points in the plane.
19. Using the even / odd identities for cosine and sine, show $A{(\theta )}^{-1}=A(-\theta )$. Interpret this geometrically.

Graph the following equations.

The matrix $$A(\theta )=\left[\begin{array}{cc}\hfill \cos (\theta )& \hfill -\sin (\theta )\\ \hfill \sin (\theta )& \hfill \cos (\theta )\end{array}\right]$$ is called a rotation matrix. We've seen this matrix most recently in the proof of used in the proof of Theorem.

Answers

1. $$x^2+2xy+y^2-x\sqrt{2}+y\sqrt{2}-6=0$$ becomes ${(x^{\prime })}^2=-(y^{\prime }-3)$ after rotating counter-clockwise through $\theta =\frac{\pi }{4}$.

   

2. $$7x^2-4xy\sqrt{3}+3y^2-2x-2y\sqrt{3}-5=0$$ becomes $\frac{{(x^{\prime }-2)}^2}{9}+{(y^{\prime })}^2=1$ after rotating counter-clockwise through $\theta =\frac{\pi }{3}$

   

3. $$5x^2+6xy+5y^2-4\sqrt{2}x+4\sqrt{2}y=0$$ becomes ${(x^{\prime })}^2+\frac{{(y^{\prime }+2)}^2}{4}=1$ after rotating counter-clockwise through $\theta =\frac{\pi }{4}$.

   

4. $$x^2+2\sqrt{3}xy+3y^2+2\sqrt{3}x-2y-16=0$$ becomes${(x^{\prime })}^2=y^{\prime }+4$ after rotating counter-clockwise through $\theta =\frac{\pi }{3}$

   

5. $$13x^2-34xy\sqrt{3}+47y^2-64=0$$ becomes ${(y^{\prime })}^2-\frac{{(x^{\prime })}^2}{16}=1$ after rotating counter-clockwise through $\theta =\frac{\pi }{6}$.

   

6. $x^2-2\sqrt{3}xy-y^2+8=0$ becomes $\frac{{(x^{\prime })}^2}{4}-\frac{{(y^{\prime })}^2}{4}=1$ after rotating counter-clockwise through $\theta =\frac{\pi }{3}$

   

7. $$x^2-4xy+4y^2-2x\sqrt{5}-y\sqrt{5}=0$$ becomes ${(y^{\prime })}^2=x$ after rotating counter-clockwise through $\theta =\arctan \left(\frac{1}{2}\right)$.

   

8. $$8x^2+12xy+17y^2-20=0$$ becomes ${(x^{\prime })}^2+\frac{{(y^{\prime })}^2}{4}=1$ after rotating counter-clockwise through $\theta =\arctan (2)$

   

9. $r=\frac{2}{1-\cos (\theta )}$ is a parabola directrix $x=-2$, vertex $(-1,0)$ focus $(0,0)$, focal diameter $4$

   

10. $r=\frac{3}{2+\sin (\theta )}=\frac{\frac{3}{2}}{1+\frac{1}{2}\sin (\theta )}$ is an ellipse directrix $y=3$, vertices $(0,1)$, $(0,-3)$ center $(0,-2)$, foci $(0,0)$, $(0,-2)$ minor axis length $2\sqrt{3}$

    

11. $r=\frac{3}{2-\cos (\theta )}=\frac{\frac{3}{2}}{1-\frac{1}{2}\cos (\theta )}$ is an ellipse directrix $x=-3$, vertices $(-1,0)$, $(3,0)$ center $(1,0)$, foci $(0,0)$, $(2,0)$ minor axis length $2\sqrt{3}$

    

12. $r=\frac{2}{1+\sin (\theta )}$ is a parabola directrix $y=2$, vertex $(0,1)$ focus $(0,0)$, focal diameter $4$

    

13. $r=\frac{4}{1+3\cos (\theta )}$ is a hyperbola directrix $x=\frac{4}{3}$, vertices $(1,0)$, $(2,0)$ center $(\frac{3}{2},0)$, foci $(0,0)$, $(3,0)$ conjugate axis length $2\sqrt{2}$

    

14. $r=\frac{2}{1-2\sin (\theta )}$ is a hyperbola directrix $y=-1$, vertices $(0,-\frac{2}{3})$, $(0,-2)$ center $(0,-\frac{4}{3})$, foci $(0,0)$, $(0,-\frac{8}{3})$ conjugate axis length $\frac{2\sqrt{3}}{3}$

    

15. $r=\frac{2}{1+\sin (\theta -\frac{\pi }{3})}$ is the parabola $r=\frac{2}{1+\sin (\theta )}$ rotated through $\varphi =\frac{\pi }{3}$

    

16. $r=\frac{6}{3-\cos \left(\theta +\frac{\pi }{4}\right)}$ is the ellipse $r=\frac{6}{3-\cos \left(\theta \right)}=\frac{2}{1-\frac{1}{3}\cos \left(\theta \right)}$ rotated through $\varphi =-\frac{\pi }{4}$