This is my contribution. Observe that all these equations are equivalent
$$x-2y-3=0\Leftrightarrow y=\frac{1}{2}x-\frac{3}{2}\Leftrightarrow \frac{x}{3}+\frac{y}{-3/2}=1.$$
And the same idea applies to the general equation $Ax+Bx+C=0$.
Lines Common point. Suppose you have two straight lines with equations
$$Ax+By+C=0$$
$$A^{\prime }x+B^{\prime }y+C^{\prime }=0.$$
They are not parallel if and only if $AB^{\prime }-A^{\prime }B\neq 0$ (or
equivalently $\frac{B^{\prime }}{B}\neq \frac{A^{\prime }}{A}$)
Since for the lines with equations
$$x-2y-3=0$$
$$3x-2y-1=0$$
we have $\frac{2}{2}\neq \frac{3}{1}$, they cross each other. To find the
coordinates $(x,y)$ of the intersecting point you have to solve the system
of equations. For instance as follows
$$\left\{
\begin{array}{c}
x-2y-3=0 \\
3x-2y-1=0%
\end{array}%
\right. \Leftrightarrow \left\{
\begin{array}{c}
x-2y-3=0 \\
-3x+2y+1=0%
\end{array}%
\right. $$
$$\Leftrightarrow \left\{
\begin{array}{c}
x-2y-3=0 \\
-2x-2=0%
\end{array}%
\right. \Leftrightarrow \left\{
\begin{array}{c}
-1-2y-3=0 \\
x=-1%
\end{array}%
\right. \Leftrightarrow \left\{
\begin{array}{c}
y=-2 \\
x=-1%
\end{array}%
\right. $$
The second system results from the first by multiplying the second equation
by $-1$. The third, by replacing the second equation by the sum of both
equations.
If you solve with matrices
$$%
\begin{bmatrix}
A & B \\
A^{\prime } & B^{\prime }%
\end{bmatrix}%
\begin{bmatrix}
x \\
y%
\end{bmatrix}%
=%
\begin{bmatrix}
-C \\
-C^{\prime }%
\end{bmatrix}%
$$
the expression $AB^{\prime }-A^{\prime }B$ is the determinant of the matrix formed by the coefficients of $x$ and $y$
$$\begin{bmatrix}
A & B \\
A^{\prime } & B^{\prime }%
\end{bmatrix}$$
Angles between lines. The smallest angle $0\leq \theta \leq \pi /2$ between two lines is such that
$$\tan \theta =\left\vert \frac{m-m^{\prime }}{1+mm^{\prime }}\right\vert ,$$
where $m$ is the slope of the line with equation $Ax+By+C=0$ and $m^{\prime }
$ is the sope of $A^{\prime }x+B^{\prime }y+C^{\prime }=0$.
This result is derived from the trigonometric identity
$$\tan (\alpha -\alpha')=\frac{\tan \alpha-\tan \alpha'}{1+\tan \alpha\cdot\tan \alpha'}$$
where $\alpha,\alpha'$ are the angles of inclination of the first and the second line, respectivelly. And $m=\tan \alpha,m'=\tan \alpha'$.
The other angle between these two lines is $\pi -\theta $.
It's hard to tell from the picture exactly what you're using as the definition of the ellipse. One way of defining an ellipse is that the sum of distances from the foci to any
point on the ellipse is constant. So let's say the foci are at $(a,0)$ and $(-a,0)$ and the sum of the distances is $d$. Then the point $(x,y)$ is on the ellipse if
$$ \sqrt{(x-a)^2 + y^2} + \sqrt{(x+a)^2 + y^2} = d $$
Move the second square root to the right side, square both sides, and expand. You get
$$ (x-a)^2 + y^2 = (x+a)^2 + y^2 - 2 d \sqrt{(x+a)^2 + y^2} + d^2 $$
After some more expansion, cancellation, and moving terms around,
$$ 2 d \sqrt{(x+a)^2 + y^2} = d^2 + 4 a x $$
Again square both sides and expand:
$$ 4\,{d}^{2}{x}^{2}+8\,{d}^{2}ax+4\,{a}^{2}{d}^{2}+4\,{d}^{2}{y}^{2}=16
\,{x}^{2}{a}^{2}+8\,{d}^{2}ax+{d}^{4}$$
Cancel the $8 d^2 a x$'s, bring the terms containing $x^2$ and $y^2$ to the left and the constant $4 a^2 d^2$ to the right, and collect terms:
$$ (4 d^2 - 16 a^2) x^2 + 4 d^2 y^2 = d^4 - 4 a^2 d^2 $$
Best Answer
$$x^2+y(y+1)^2=0$$ gives something like what you want.
See also Graph of an infinitely extending rollercoaster loop