Asymptotes to Hyperbola in Reduced Form

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Theorem

Let $\KK$ be a hyperbola embedded in a cartesian plane in reduced form with the equation:

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


$\KK$ has two asymptotes which can be described by the equation:

$\dfrac {x^2} {a^2} - \dfrac {y^2} {b^2} = 0$

that is:

$y = \pm \dfrac b a x$


Proof

From Equation of Hyperbola in Reduced Form: Polar Frame, $\KK$ can be described in polar coordinates as:

$\dfrac {\cos^2 \theta} {a^2} - \dfrac {\sin^2 \theta} {b^2} = \dfrac 1 {r^2}$

When $\theta = 0$ we have that $r = a$.

As $\theta$ increases, $\cos^2 \theta$ decreases and $\sin^2 \theta$ increases.

Hence $\dfrac 1 {r^2}$ decreases and $r$ increases as a consequence.

This continues until:

$\dfrac {\cos^2 \theta} {a^2} - \dfrac {\sin^2 \theta} {b^2} = 0$

that is:

\(\ds \tan^2 \theta\) \(=\) \(\ds \dfrac {b^2} {a^2}\)
\(\ds \dfrac {y^2} {x^2}\) \(=\) \(\ds \dfrac {b^2} {a^2}\)
\(\ds \dfrac {x^2} {a^2} - \dfrac {y^2} {b^2}\) \(=\) \(\ds 0\)

This can be expressed in the form:

$b^2 x^2 - a^2 y^2 = 0$

This is a homogeneous quadratic equation in $2$ variables of the form:

$a' x^2 + 2 h' x y + b' y^2 0 $

where:

$h' = 0$
$a' = b^2$
$b' = a^2$

From Homogeneous Quadratic Equation represents Two Straight Lines through Origin, this is the equation for $2$ straight lines through the origin:

$y = \pm \dfrac b a x$

$\blacksquare$


Sources