Equation of Hyperbola in Reduced Form/Cartesian Frame

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Theorem

Let $K$ be a hyperbola such that:

the transverse axis of $K$ has length $2 a$
the conjugate axis of $K$ has length $2 b$.


Let $K$ be aligned in a cartesian plane in reduced form.


$K$ can be expressed by the equation:

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


Parametric Form 1

The right-hand branch of $K$ can be expressed in parametric form as:

$\begin {cases} x = a \cosh \theta \\ y = b \sinh \theta \end {cases}$


Parametric Form 2

$K$ can be expressed in parametric form as:

$\begin {cases} x = a \sec \theta \\ y = b \tan \theta \end {cases}$


Proof 1

 

HyperbolaReducedForm.png


By definition, the foci $F_1$ and $F_2$ of $K$ are located at $\tuple {-c, 0}$ and $\tuple {c, 0}$ respectively.

Let the vertices of $K$ be $V_1$ and $V_2$.

By definition, these are located at $\tuple {-a, 0}$ and $\tuple {a, 0}$.

Let the covertices of $K$ be $C_1$ and $C_2$.

By definition, these are located at $\tuple {0, -b}$ and $\tuple {0, b}$.


Let $P = \tuple {x, y}$ be an arbitrary point on the locus of $K$.


From the equidistance property of $K$ we have that:

$\size {F_1 P - F_2 P} = d$

where $d$ is a constant for this particular ellipse.

From Equidistance of Hyperbola equals Transverse Axis:

$d = 2 a$

Also, from Focus of Hyperbola from Transverse and Conjugate Axis:

$c^2 a^2 = b^2$

Without loss of generality, let us choose a point $P$ such that $F_1 P > F_2 P$.

Then:

\(\ds \sqrt {\paren {x + c}^2 + y^2} - \sqrt {\paren {x - c}^2 + y^2}\) \(=\) \(\ds d = 2 a\) Pythagoras's Theorem
\(\ds \leadsto \ \ \) \(\ds \sqrt {\paren {x + c}^2 + y^2}\) \(=\) \(\ds 2 a + \sqrt {\paren {x - c}^2 + y^2}\)
\(\ds \leadsto \ \ \) \(\ds \paren {x + c}^2 + y^2\) \(=\) \(\ds \paren {2 a + \sqrt {\paren {x - c}^2 + y^2} }^2\) squaring both sides
\(\ds \leadsto \ \ \) \(\ds x^2 + 2 c x + c^2 + y^2\) \(=\) \(\ds 4 a^2 + 4 a \sqrt {\paren {x - c}^2 + y^2} + \paren {x - c}^2 + y^2\) expanding
\(\ds \leadsto \ \ \) \(\ds x^2 + 2 c x + c^2 + y^2\) \(=\) \(\ds 4 a^2 + 4 a \sqrt {\paren {x - c}^2 + y^2} + x^2 - 2 c x + c^2 + y^2\) further expanding
\(\ds \leadsto \ \ \) \(\ds c x - a^2\) \(=\) \(\ds a \sqrt {\paren {x - c}^2 + y^2}\) gathering terms and simplifying
\(\ds \leadsto \ \ \) \(\ds \paren {c x - a^2}^2\) \(=\) \(\ds a^2 \paren {\paren {x - c}^2 + y^2}^2\) squaring both sides
\(\ds \leadsto \ \ \) \(\ds c^2 x^2 - 2 c x a^2 + a^4\) \(=\) \(\ds a^2 x^2 - 2 c x a^2 + a^2 c^2 + a^2 y^2\) expanding
\(\ds \leadsto \ \ \) \(\ds c^2 x^2 + a^4\) \(=\) \(\ds a^2 x^2 + a^2 c^2 + a^2 y^2\) simplifying
\(\ds \leadsto \ \ \) \(\ds a^2 c^2 - a^4\) \(=\) \(\ds c^2 x^2 - a^2 x^2 - a^2 y^2\) gathering terms
\(\ds \leadsto \ \ \) \(\ds a^2 \paren {c^2 - a^2}\) \(=\) \(\ds \paren {c^2 - a^2} x^2 - a^2 y^2\) simplifying
\(\ds \leadsto \ \ \) \(\ds a^2 b^2\) \(=\) \(\ds b^2 x^2 - a^2 y^2\) substituting $c^2 - a^2 = b^2$ from $(2)$
\(\ds \leadsto \ \ \) \(\ds 1\) \(=\) \(\ds \frac {x^2} {a^2} - \frac {y^2} {b^2}\) dividing by $a^2 b^2$

$\blacksquare$


Proof 2


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