Equation of Confocal Hyperbolas/Formulation 1
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Definition
The equation:
- $(1): \quad \dfrac {x^2} {a^2 + \lambda} + \dfrac {y^2} {b^2 + \lambda} = 1$
where:
- $\tuple {x, y}$ denotes an arbitrary point in the cartesian plane
- $a$ and $b$ are (strictly) positive constants such that $a^2 > b^2$
- $\lambda$ is a (strictly) positive parameter such that $b^2 < -\lambda < a^2$
defines the set of all confocal hyperbolas whose foci are at $\tuple {\pm \sqrt {a^2 + b^2}, 0}$.
Proof
Let $a$ and $b$ be arbitrary (strictly) positive real numbers fulfilling the constraints as defined.
Let $E$ be the locus of the equation:
- $(1): \quad \dfrac {x^2} {a^2 + \lambda} + \dfrac {y^2} {b^2 + \lambda} = 1$
As $b^2 < -\lambda$ it follows that:
- $b^2 + \lambda < 0$
and as $-\lambda < a^2$:
- $a^2 + \lambda > 0$
Thus $(1)$ is in the form:
- $\dfrac {x^2} {r^2} - \dfrac {y^2} {s^2} = 1$
where:
- $r^2 = a^2 + \lambda$
- $s^2 = -\lambda + b^2$
From Equation of Hyperbola in Reduced Form, this is the equation of an hyperbola in reduced form.
It follows that:
- $\tuple {\pm \sqrt {a^2 + \lambda}, 0}$ are the positions of the vertices of $E$
- $\tuple {0, \pm \sqrt {b^2 - \lambda} }$ are the positions of the covertices of $E$
From Focus of Hyperbola from Transverse and Conjugate Axis, the positions of the foci of $E$ are given by:
- $\paren {a^2 + \lambda} + \paren {b^2 - \lambda} = c^2$
where $\tuple {\pm c, 0}$ are the positions of the foci of $E$.
Thus we have:
\(\ds c^2\) | \(=\) | \(\ds \paren {a^2 + \lambda} + \paren {b^2 - \lambda}\) | Focus of Hyperbola from Transverse and Conjugate Axis | |||||||||||
\(\ds \) | \(=\) | \(\ds a^2 + b^2\) |
Hence the result.
$\blacksquare$