# Equation of Confocal Hyperbolas/Formulation 1

## 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:

 $\displaystyle c^2$ $=$ $\displaystyle \paren {a^2 + \lambda} + \paren {b^2 - \lambda}$ Focus of Hyperbola from Transverse and Conjugate Axis $\displaystyle$ $=$ $\displaystyle a^2 + b^2$

Hence the result.

$\blacksquare$