derivation of equation of a hyperbola from the conic section This is a method I tried of working out the cartesian form of a hyperbola using its definition as the cross section of a vertical cone?
I started by noting that the width of the hyperbola is the sine of a circle at some point within the cone. The radius of the circle is proportional to the height of the cone, which is the length $y$ down the hyperbola plus some constant $a$. Therefore the radius of the circle is $k(y+a)$.
The centre of the hyperbola is always the same distance from the centre of the cone at a given height $y$, at $ka$. So cosine of the angle between the middle and edge of the hyperbola at some height $y$ is $\frac{ka}{k(y+a)} = \frac{1}{1+\frac{y}{a}}$. So the width of the hyperbola $x$ at height $y$ is $x =k(y+a)\sqrt{1-\frac{1}{(1+\frac{y}{a})^2}}$ by relating the sines to cosines. The resulting equation is $x^2 = k^2(y+a)^2(1-\frac{1}{(1+\frac{y}{a})^2})$,
This simplifies to $x^2 = k^2(y^2 + 2ya)$. Is this a hyperbolic curve. How do I rearrange it to the standard form?
I understand that I'm not being very clear but I can't find a software on which to draw the problem properly. Anyway, I hope that the line of reasoning has come across.
 A: $$
\eqalign{
  & x^{\,2}  = k^{\,2} \left( {y + a} \right)^{\,2} \left( {1 - {1 \over {\left( {1 + y/a} \right)^{\,2} }}} \right) = k^{\,2} a^{\,2} \left( {1 + y/a} \right)^{\,2} \left( {{{\left( {1 + y/a} \right)^{\,2}  - 1} \over {\left( {1 + y/a} \right)^{\,2} }}} \right) =   \cr 
  &  = k^{\,2} a^{\,2} \left( {\left( {1 + y/a} \right)^{\,2}  - 1} \right) = k^{\,2} \left( {\left( {y + a} \right)^{\,2}  - a^{\,2} } \right) \cr} 
$$
that is
$$
{{\left( {y + a} \right)^{\,2} } \over {a^{\,2} }} - {{x^{\,2} } \over {\left( {ka} \right)^{\,2} }} = 1
$$
which looks correct in the reference system you adopted:
$$
\left\{ \matrix{
  x = 0\quad  \to \quad y =  - a \pm a \hfill \cr 
  x =  \pm ka\quad  \to \quad y =  - a \pm \sqrt 2 a \hfill \cr}  \right.
$$
A: $-\frac{x^2}{k^2} + (y+a)^2 = a^2$
$-\frac{x^2}{a^2k^2} + \frac{(y+a)^2}{a^2} = 1$
This is the equation of a hyperbola on the vertical transverse axis with vertex $a$, asymptote, $y=\frac{1}{k}x$ and translated down by $a$, indeed a correct parametrisation for the curve of the hyperbola
