An ellipse in polar coordinates

Let $A\cos(\theta+\phi)=3\cos\theta+2\sin\theta$ . Expanding and solving gives $A=\sqrt{13}$ and $\phi=-\arccos\frac3{\sqrt{13}}$ , so $r=\frac1{5+\sqrt{13}\cos\left(\theta-\phi\right)}$ Let's rotate the entire ellipse by removing $\phi$ to simplify this equation. This does not affect the shape of the ellipse. $r=\frac1{5+\sqrt{13}\cos(\theta)}$ $5r+\sqrt{13}r\cos(\theta)=1$ $25r^2=\left(1-\sqrt{13}r\cos(\theta)\right)^2$ Now we convert to Cartesian coordinates with $r^2=x^2+y^2$ and $r\cos\theta=x$ $25x^2+25y^2=\left(1-\sqrt{13}x\right)^2$ Expanding and completing the square for the $x$ terms gives $12\left(x+\frac{\sqrt{13}}{12}\right)^2+25y^2=\frac{25}{12}$ Let's translate the entire ellipse in the $x$ direction by removing $\frac{\sqrt{13}}{12}$ . This does not affect the shape of the ellipse. $12x^2+25y^2=\frac{25}{12}$ $\frac{144}{25}x^2+12y^2=1$ This is the equation of an ellipse with semi-major axis $\sqrt{\frac{25}{144}}=\frac5{12}$ and semi-minor axis $\frac1{\sqrt{12}}$ , so the answer is $\frac5{12}+\frac1{\sqrt{12}}\approx0.7053$

Since $r = \frac{1}{5 + \sqrt{13}\cos(\theta + \alpha)}$ for some angle $\alpha$ , we see that $\tfrac1{12}(5-\sqrt{13}) = (5 +\sqrt{13})^{-1} \le r \le (5 - \sqrt{13})^{-1} = \tfrac{1}{12}(5 + \sqrt{13})$ . Since the maximum and minimum distances from the focus of an elliptical orbit are $a(1+e)$ and $a(1-e)$ , we deduce that $a = \tfrac{5}{12}$ and $ae = \tfrac{1}{12}\sqrt{13}$ . Thus $e = \tfrac{\sqrt{13}}{5}$ and hence $b = \tfrac1{\sqrt{12}}$ , so $a+b=\tfrac{1}{12}(5 + \sqrt{12}) = \boxed{0.7053418013}$ .

Polar equation of an ellipse is

$r(\theta) =\dfrac {l}{1+e\cos \theta}$ ,

where $l=\dfrac {b^2}{a}=$ semi latus rectum,

$e=\sqrt {1-\frac{b^2}{a^2}}=$ eccentricity,

$a, b$ are the semi-major and the semi-minor axes of the ellipse respectively.

Here

$r(\theta) =\dfrac {1}{5+3\cos \theta+2\sin \theta}$

$=\dfrac {\frac 15}{1+\frac{\sqrt {13}}{5}\cos (\theta-\tan^{-1} \frac 23)}$

So, $l=\dfrac {b^2}{a}=\dfrac 15$ ,

$e=\sqrt {1-\frac{b^2}{a^2}}=\dfrac {\sqrt {13}}{5}$

Solving we get $a=\dfrac {5}{12}$ ,

$b=\dfrac {1}{\sqrt {12}}$ , and

$a+b=\dfrac {5+\sqrt {12}}{12}\approx$

$\boxed {7053418}$

We're given that,

$r = \dfrac{1}{a + b \cos \theta + c \sin \theta}$

From the definition of polar coordinates, we have the polar-to-rectangular coordinates relation

$x = r \cos \theta, y = r \sin \theta \hspace{20pt} (1)$ .

First, we'll eliminate having to deal with three constants a, b, c , by combining $b \cos \theta$ and $c \sin \theta$ into one $\theta$ -shifted sinusoid. If $\phi = \text{Arctan2}( b , c )$ , then,

$r = \dfrac{1}{a + b' \cos(\theta - \phi) }$

where $b' = \sqrt{ b^2 + c^2 }$ . Rotating our $x$ -axis to $\theta = \phi$ , the equation becomes,

$r = \dfrac{1}{ a + b' \cos \theta } \hspace{20pt} (2)$

In the following, we will replace $b'$ with $b$ , for simplicity of notation.

Using $(1)$ and $(2)$ , we get,

$x (a + b \cos \theta ) = \cos \theta \hspace{20pt} (3.1)$

$y (a + b \cos \theta ) = \sin \theta \hspace{20pt} (3.2)$

Equations $(3)$ are a linear system in $\cos \theta$ and $\sin \theta$ , and if we define

$u = \begin{bmatrix} \cos \theta \\ \sin \theta \end{bmatrix}$

then, this linear system can be written as $A u = B$ , where

$A = \begin{bmatrix} (x b - 1) && 0 \\ y b && -1 \end{bmatrix}$

and

$B = \begin{bmatrix} - xa \\ -y a \end{bmatrix}$

Solving, we get,

$\cos \theta = \dfrac{ ax }{1 - b x }$ and $\sin \theta = \dfrac{ a y }{1 - b x}$

Now since $\cos^2 \theta + \sin^2 \theta = 1$ , then,

$(a x)^2 + (a y)^2 = (1 - b x )^2$

$\rightarrow a^2 ( x^2 + y^2 ) = 1 + b^2 x^2 - 2 b x$

$\rightarrow (a^2 - b^2) x^2 + a^2 y^2 - 2 b x = 1$

Completing the square in $x$ , we obtain,

$(a^2 - b^2) (x - \dfrac{b}{a^2 - b^2} )^2 + a^2 y = 1 + \dfrac{b^2}{a^2 - b^2} = \dfrac{a^2}{a^2 - b^2}$

Dividing by the right hand side,

$\dfrac{(a^2 - b^2)^2}{a^2} (x - x_0)^2 + (a^2 - b^2) y^2 = 1 \hspace{20pt} (4)$

where $x_0 = \dfrac{b}{a^2 - b^2}$

From equation (4) , it is evident that the semi-major axis is given by $\dfrac{a}{a^2 - b^2}$ and the semi-minor axis is given by $\dfrac{1}{\sqrt{a^2 - b^2}}$ .

In this problem $a = 5 , b = 3 , c = 2$ , so our $b' = \sqrt{13}$ which we replaced with $b$ .

Hence the semi-major axis $= \dfrac{5}{25 - 13} = \dfrac{5}{12}$ , and the semi-minor axis is $\dfrac{1}{\sqrt{12}}$ . Their sum comes to $\approx 0.7053$ .