avewright/proofwiki-math huggingface dataset:
Theorem: Time of Travel down Brachistochrone

Let a wire $AB$ be curved into the shape of a brachistochrone.
Let $AB$ be embedded in a constant and uniform gravitational field where Acceleration Due to Gravity is $g$.
Let a bead $P$ be released at $A$ to slide down without friction to $B$.
Then the time taken for $P$ to slide from $A$ to $B$ is:
:$T = \pi \sqrt {\dfrac a g}$
where $a$ is the radius of the generating circle of the cycloid which forms $AB$.

Proof:
That the curve $AB$ is indeed a cycloid is demonstrated in Brachistochrone is Cycloid.
Let $A$ be located at the origin of a cartesian plane.
We have the equations of the cycloid:
{{begin-eqn}}
{{eqn | l = x
      | r = a \paren {\theta - \sin \theta}
}}
{{eqn | l = y
      | r = a \paren {1 - \cos \theta}
}}
{{end-eqn}}
Differentiating with respect to theta:
{{begin-eqn}}
{{eqn | l = \frac {\d x} {\d \theta}
      | r = \paren {1 - \cos \theta}
      | c = 
}}
{{eqn | l = \frac {\d y} {\d \theta}
      | r = a \sin \theta
}}
{{end-eqn}}
Let $s$ be the distance along $AB$ from the origin.
Then:
{{begin-eqn}}
{{eqn | l = \paren {\frac {\d s} {\d \theta} }^2
      | r = \paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2
      | c = 
}}
{{eqn | r = a^2 \paren {1 - \cos \theta}^2 + a^2 \sin^2 \theta
      | c = 
}}
{{eqn | r = a^2 \paren {1 - 2 \cos \theta + \cos^2 \theta + \sin^2 \theta}
      | c = 
}}
{{eqn | r = 2 a^2 \paren {1 - \cos \theta}
      | c = 
}}
{{eqn | r = 4 a^2 \map {\sin^2} {\frac \theta 2}
      | c = 
}}
{{end-eqn}}
Thus:
:$\dfrac {\d s} {\d \theta} = 2 a \map \sin {\dfrac \theta 2}$
Now:
{{begin-eqn}}
{{eqn | l = v
      | r = \dfrac {\d s} {\d t}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \rd t
      | r = t
      | c = 
}}
{{eqn | r = \int \frac {\d s} v
      | c = 
}}
{{end-eqn}}
We also have, from the Principle of Conservation of Energy, that:
{{begin-eqn}}
{{eqn | l = m g y
      | r = \frac {m v^2} 2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \dfrac {\d s} {\d t}
      | c = 
}}
{{eqn | r = \sqrt {2 g y}
      | c = 
}}
{{eqn | r = \sqrt {2 g a \paren {1 - \cos \theta} }
      | c = 
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \frac {\d t} {\d \theta}
      | r = \frac {\d t} {\d s} \theta
      | c = 
}}
{{eqn | r = \frac {a \sqrt {2 \paren {1 - \cos \theta} } } {\sqrt {2 g a \paren {1 - \cos \theta} } }
      | c = 
}}
{{eqn | r = \sqrt {\frac a g}
      | c = 
}}
{{end-eqn}}
Let $T$ be the time taken to slide down the brachistochrone.
At the top:
:$\theta = 0$
and at the bottom:
:$\theta = \pi$
Then:
{{begin-eqn}}
{{eqn | l = T
      | r = \int \rd t
      | c = 
}}
{{eqn | r = \sqrt {\frac a g} \int_0^\pi \rd \theta
      | c = 
}}
{{eqn | r = \pi \sqrt {\frac a g}
      | c = 
}}
{{end-eqn}}
So the time to slide down this brachistochrone from the top to the bottom is:
:$T = \pi \sqrt {\dfrac a g}$
{{qed}}
Theorem: Time of Travel down Brachistochrone/Corollary

Let a bead $P$ be released from anywhere on the wire between $A$ and $B$ to slide down without friction to $B$.
Then the time taken for $P$ to slide to $B$ is:
:$T = \pi \sqrt{\dfrac a g}$

Proof:
That the curve $AB$ is indeed a cycloid is demonstrated in Brachistochrone is Cycloid.
Let $A$ be located at the origin of a cartesian plane.
Let the bead slide from an intermediate point $\theta_0$.
We have:
:$v = \dfrac {\d s} {\d t} = \sqrt {2 g \paren {y - y_0} }$
which leads us, via the same route as for Time of Travel down Brachistochrone, to:
{{begin-eqn}}
{{eqn | l = T
      | r = \int_{\theta_0}^\pi \sqrt {\frac {2 a^2 \paren {1 - \cos \theta} } {2 g a \paren {\cos \theta_0 - \cos \theta} } } \rd \theta
      | c = 
}}
{{eqn | r = \sqrt {\frac a g} \int_{\theta_0}^\pi \sqrt {\frac {1 - \cos \theta} {\cos \theta_0 - \cos \theta} } \rd \theta
      | c = 
}}
{{end-eqn}}
Using the Half Angle Formula for Cosine and Half Angle Formula for Sine, this gives:
:$\ds T = \sqrt {\frac a g} \int_{\theta_0}^\pi \frac {\map \sin {\theta / 2} } {\sqrt {\map \cos {\theta_0 / 2} - \map \cos {\theta / 2} } } \rd \theta$
Now we make the substitution:
{{begin-eqn}}
{{eqn | l = u
      | r = \frac {\map \cos {\theta / 2} } {\map \cos {\theta_0 / 2} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d \theta}
      | r = -\frac {\map \sin {\theta / 2} } {2 \map \cos {\theta_0 / 2} }
      | c = 
}}
{{end-eqn}}
Recalculating the limits:
:when $\theta = \theta_0$ we have $u = 1$
:when $\theta = \pi$ we have $u = 0$.
So:
{{begin-eqn}}
{{eqn | l = T
      | r = -2 \sqrt {\frac a g} \int_1^0 \frac {\d u} {\sqrt {1 - u^2} }
      | c = 
}}
{{eqn | r = \intlimits {2 \sqrt {\frac a g} \sin^{-1} u} 0 1
      | c = 
}}
{{eqn | r = \pi \sqrt {\frac a g}
      | c = 
}}
{{end-eqn}}
Thus the time to slide down a brachistochrone from any arbitrary point $\theta_0$ is:
:$T = \pi \sqrt {\dfrac a g}$
{{qed}}
Theorem: Rational Numbers are F-Sigma Set in Real Line

Let $\struct {\R, \tau}$ be the real number line considered asa  topological space with the usual (Euclidean) topology.
Then:
:$\Q$ is an $F_\sigma$ set in $\struct {\R, \tau}$.

Proof:
Define the set of subsets of $\R$ as:
:$\FF := \set {\set x: x \in \Q}$
By Closed Real Interval is Closed Set:
:$\forall x \in \Q: \closedint x x = \set x$ is closed (in topological sense)
Then:
:$\forall A \in \FF: A$ is closed
By Cardinality of Set of Singletons:
:$\card \FF = \card \Q$
where $\card \FF$ denotes the cardinality of $\FF$.
By Rational Numbers are Countably Infinite:
:$\Q$ is countable.
Therefore by Set is Countable if Cardinality equals Cardinality of Countable Set:
:$\FF$ is countable.
By Union of Set of Singletons:
:$\bigcup \FF = \Q$
Thus, by definition, $\Q$ is an $F_\sigma$ set.
{{qed}}
Theorem: Union of Set of Singletons

Let $S$ be a set.
Let $T = \set {\set x: x \in S}$ be the set of all singletons of elements of $S$.
Then:
:$\ds \bigcup T = S$
where $\ds \bigcup T$ denotes the union of $T$.

Proof:
=== Union of $T$ Subset $S$ ===
Let $\ds x \in \bigcup T$.
By definition of union:
:$\exists A \in T: x \in A$
By definition of $T$:
:$\exists y \in S: A = \set y$
Then by definition of singleton:
:$x = y$
Thus $x \in S$.
{{qed|lemma}}
Theorem: Confocal Conics are Self-Orthogonal

The confocal conics defined by:
:$\quad \dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$
forms a family of orthogonal trajectories which is self-orthogonal.
:500px

Proof:
Consider:
:$(1): \quad \dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$
From Equation of Confocal Ellipses: Formulation 2:
:$(1)$ defines an ellipse when $a^2 > c^2$.
From Equation of Confocal Hyperbolas: Formulation 2:
:$(1)$ defines a hyperbola when $a^2 < c^2$.
Thus it is seen that $(1)$ is that of a conic section.
We use the technique of formation of ordinary differential equation by elimination.
Differentiating {{WRT|Differentiation}} $x$ gives:
:$\dfrac {2 x} {a^2} + \dfrac {2 y} {a^2 - c^2} \dfrac {\d y} {\d x} = 0$
so
:$\dfrac {\d y} {\d x} = - \dfrac {a^2 - c^2} {a^2} \dfrac x y$
Now we need to eliminate $c$ from the above.
We go back to $(1)$:
{{begin-eqn}}
{{eqn | l = \frac {x^2} {a^2} + \frac {y^2} {a^2 - c^2}
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {y^2} {a^2 - c^2}
      | r = 1 - \frac {x^2} {a^2}
      | c = 
}}
{{eqn | r = \frac {a^2 - x^2}{a^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a^2 - c^2
      | r = \frac {a^2 y^2} {a^2 - x^2}
      | c = 
}}
{{end-eqn}}
Substituting for $a^2 - c^2$:
{{begin-eqn}}
{{eqn | l = \dfrac {\d y} {\d x}
      | r = -\frac {a^2 y^2} {a^2 \paren {a^2 - x^2} } \frac x y
      | c = 
}}
{{eqn | r = \frac {x y} {a^2 - x^2}
      | c = 
}}
{{end-eqn}}
This is separable, so separate it:
:$\ds \int \frac {\d y} y = -\int \frac {x \rd x} {a^2 - x^2}$
which leads to:
{{begin-eqn}}
{{eqn | l = \ln y
      | r = \frac 1 2 \, \map \ln {a^2 - x^2} + \ln k
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 \ln y
      | r = \map \ln {a^2 - x^2} + 2 \ln k
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \ln y^2
      | r = \map \ln {a^2 - x^2} + \ln k^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \ln y^2
      | r = \map \ln {k^2 \paren {a^2 - x^2} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y^2
      | r = k^2 \paren {a^2 - x^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y^2 + k^2 x^2
      | r = a^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {x^2} {a^2} + \frac {y^2} {k^2 a^2}
      | r = 1
      | c = 
}}
{{end-eqn}}
Now $k^2$ is still arbitrary at this point, so set:
:$k^2 = \dfrac {a^2 - c^2} {a^2}$
Substituting this into the above gives us:
:$\dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$
which is $(1)$.
Hence the result by definition of self-orthogonal.
{{qed}}
Theorem: T1 Space is T1/2 Space

Let $T$ be a $T_1$ topological space.
Then $T$ is $T_{\frac 1 2}$ space.

Proof:
By Closure of Derivative is Derivative in T1 Space:
:$\forall A \subseteq T: \paren {A'}^- = A'$
where
:$A'$ denotes the derivative of $A$
:$\paren {A'}^-$ denotes the closure of $A'$
Then by Topological Closure is Closed:
:$\forall A \subseteq T: A'$ is closed
Thus by definition:
:$T$ is $T_{\frac 1 2}$ space
{{qed}}
Theorem: T1/2 Space is T0 Space

Let $T = \struct {S, \tau}$ be a $T_{\frac 1 2}$ topological space.
Then $T$ is $T_0$ space.

Proof:
By Characterization of T0 Space by Closures of Singletons it suffices to prove that
:$\forall x, y \in S: x \ne y \implies x \notin \set y^- \lor y \notin \set x^-$
where $\set y^-$ denotes the closure of $\set y$.
Let $x, y$ be points of $T$ such that:
:$x \ne y$
{{AimForCont}}:
:$x \in \set y^- \land y \in \set x^-$
We will prove that:
:$x \notin \set x'$
where $\set x'$ denotes the derivative of $\set x$.
{{AimForCont}}:
:$x \in \set x'$
As $S$ is open by definition of topological space
by Characterization of Derivative by Open Sets:
:$\exists z \in S: z \in \set x \cap S \land z \ne x$
By definition of intersection:
:$z \in \set x$
This contradicts $z \ne x$ by definition of singleton.
Thus $x \notin \set x'$.
We will prove that:
:$(1): \quad \lnot \forall G \in \tau: y \in G \implies \set x \cap G \ne \O$
{{AimForCont}}:
:$\forall G \in \tau: y \in G \implies \set x \cap G \ne \O$
As sublemma we will show that:
:$\forall U \in \tau: y \in U \implies \exists r \in S: r \in \set x \cap U \land y \ne r$
Let $U \in \tau$ such that:
:$y \in U$
Then by assumption:
:$\set x \cap U \ne \O$
By definition of empty set:
:$\exists z: z \in \set x \cap U$
By definition of intersection:
:$z \in \set x$
Then by definition of singleton:
:$z = x \ne y$
Thus:
:$\exists r \in S: r \in \set x \cap U \land y \ne r$
Then by Characterization of Derivative by Open Sets:
:$y \in \set x'$
By definition of relative complement:
:$y \notin \relcomp S {\set x'} \land x \in \relcomp S {\set x'}$
By definition of $T_{\frac 1 2}$ space:
:$\set x'$ is closed
By definition of closed set:
:$\relcomp S {\set x'}$ is open
By $x \in \set y^-$ and by Condition for Point being in Closure:
:$\set y \cap \relcomp S {\set x'} \ne \O$
Then by definition of empty set:
:$\exists z: z \in \set y \cap \relcomp S {\set x'}$
By definition of intersection:
:$z \in \set y \land z \in \relcomp S {\set x'}$
By definition of singleton:
:$z = y$
Thus $z \in \relcomp S {\set x'}$ contradicts $y \notin \relcomp S {\set x'}$.
This ends the proof of $(1)$.
Then by $(1)$ and Condition for Point being in Closure:
:$y \notin \set x^-$
This contradicts $y \in \set x^-$.
{{qed}}
Theorem: Characterization of T0 Space by Closures of Singletons

Let $T = \struct {S, \tau}$ be a topological space.
Then
:$T$ is a $T_0$ space {{iff}}:
::$\forall x, y \in S: x \ne y \implies x \notin \set y^- \lor y \notin \set x^-$
where $\set y^-$ denotes the closure of $\set y$.

Proof:
=== Sufficient Condition ===
Let $T$ be a $T_0$ space.
Let $x, y \in S$ such that
:$x \ne y$
{{AimForCont}}
:$x \in \set y^- \land y \in \set x^-$
Then:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = \set y^-
}}
{{eqn | lo= \land
      | l = \set y
      | o = \subseteq
      | r = \set x^-
}}
{{eqn | ll= \leadsto
      | l = \set x
      | o = \subseteq
      | r = \set y^-
      | c = {{Defof|Singleton}}
}}
{{eqn | lo= \land
      | l = \set y
      | o = \subseteq
      | r = \set x^-
}}
{{eqn | ll= \leadsto
      | l = \set x^-
      | o = \subseteq
      | r = \paren {\set y^-}^-
      | c = Topological Closure of Subset is Subset of Topological Closure
}}
{{eqn | lo= \land
      | l = \set y^-
      | o = \subseteq
      | r = \paren {\set x^-}^-
}}
{{eqn | ll= \leadsto
      | l = \set x^-
      | o = \subseteq
      | r = \set y^-
      | c = Closure of Topological Closure equals Closure
}}
{{eqn | lo= \land
      | l = \set y^-
      | o = \subseteq
      | r = \set x^-
}}
{{eqn | ll= \leadsto
      | l = \set x^-
      | r = \set y^-
      | c = {{Defof|Set Equality|index = 2}}
}}
{{end-eqn}}
By Characterization of $T_0$ Space by Distinct Closures of Singletons:
:$\set x^- \ne \set y^-$
This contradicts the equality.
Thus the result by Proof by Contradiction
{{qed|lemma}}
Theorem: Characterization of T0 Space by Distinct Closures of Singletons

Let $T = \struct {S, \tau}$ be a topological space.
Then
:$T$ is a $T_0$ space {{iff}}
::$\forall x, y \in S: x \ne y \implies \set x^- \ne \set y^-$
where $\set y^-$ denotes the closure of $\set y$.

Proof:
=== Sufficient Condition ===
Let $T$ be a $T_0$ space.
Let $x, y \in S$ such that
:$x \ne y$
By definition of $T_0$ space:
:$\paren {\exists U \in \tau: x \in U \land y \notin U} \lor \paren {\exists U \in \tau: x \notin U \land y \in U}$
{{WLOG}}, suppose
:$\exists U \in \tau: x \in U \land y \notin U$
By definition of singleton:
:$x \in \set x$
By Set is Subset of its Topological Closure:
:$\set x \subseteq \set x^-$
Then by definition of subset:
:$x \in \set x^-$
Hence by definition of intersection:
:$\set x^- \cap U \ne \O$
By definition of singleton:
:$\forall z: z \in \set y \implies z = y$
Then by definition of intersection:
:$\set y \cap U = \O$
Hence by Open Set Disjoint from Set is Disjoint from Closure
:$\set y^- \cap U = \O$
Thus:
:$\set x^- \ne \set y^-$
{{qed|lemma}}
Theorem: Characterization of T0 Space by Closed Sets

Let $T = \struct {S, \tau}$ be a topological space.
Then
:$T$ is a $T_0$ space {{iff}}
::for every points $x, y \in S$ if $x \ne y$ then
:::there exists a closed subset $F$ of $S$ such that $x \in F$ and $y \notin F$
::or
:::there exists a closed subset $F$ of $S$ such that $x \notin F$ and $y \in F$

Proof:
=== Sufficient Condition ===
Let $T$ be a $T_0$ space.
Let $x, y \in S$ such that
:$x \ne y$
By definition of $T_0$ space:
:$\paren {\exists U \in \tau: x \in U \land y \notin U} \lor \paren {\exists U \in \tau: x \notin U \land y \in U}$
{{WLOG}}, suppose:
:$\exists U \in \tau: x \in U \land y \notin U$
By definition:
:$\relcomp S U$ is closed
where $\relcomp S U$ denotes the relative complement of $U$ in $S$.
By definition of relative complement:
:$x \notin \relcomp S U \land y \in \relcomp S U$
Thus:
:$\exists F \subseteq S: F$ is closed $\land\, x \notin F \land y \in F$
{{qed|lemma}}
Theorem: Linear Eccentricity of Ellipse from Major and Minor Axis

Let $K$ be an ellipse whose major axis is $2 a$ and whose minor axis is $2 b$.
Let $c$ be the linear eccentricity of $K$.
Then:
:$a^2 = b^2 + c^2$

Proof:
:500px
Let the foci of $K$ be $F_1$ and $F_2$.
Let the vertices of $K$ be $V_1$ and $V_2$.
Let the covertices of $K$ be $C_1$ and $C_2$.
Let $P = \tuple {x, y}$ be an arbitrary point on the locus of $K$.
From the equidistance property of $K$ we have that:
:$F_1 P + F_2 P = d$
where $d$ is a constant for this particular ellipse.
This is true for all points on $K$.
In particular, it holds true for $V_2$, for example.
Thus:
{{begin-eqn}}
{{eqn | l = d
      | r = F_1 V_2 + F_2 V_2
      | c = 
}}
{{eqn | r = \paren {a + c} + \paren {a - c}
      | c = 
}}
{{eqn | r = 2 a
      | c = 
}}
{{end-eqn}}
It also holds true for $C_2$:
:$F_1 C_2 + F_2 C_2 = d$
Then:
{{begin-eqn}}
{{eqn | l = F_1 C_2^2
      | r = O F_1^2 + O C_2^2
      | c = Pythagoras's Theorem
}}
{{eqn | r = c^2 + b^2
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = F_1 C_2^2
      | r = O F_1^2 + O C_2^2
      | c = Pythagoras's Theorem
}}
{{eqn | r = c^2 + b^2
      | c = 
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = F_1 C_2 + F_2 C_2
      | r = 2 \sqrt {b^2 + c^2}
      | c = 
}}
{{eqn | r = 2 a
      | c = as $2 a = d$
}}
{{eqn | ll= \leadsto
      | l = a
      | r = \sqrt {b^2 + c^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | r = a^2
      | l = b^2 + c^2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Equidistance of Ellipse equals Major Axis

Let $K$ be an ellipse whose foci are $F_1$ and $F_2$.
Let $P$ be an arbitrary point on $K$.
Let $d$ be the constant distance such that:
:$d_1 + d_2 = d$
where:
:$d_1 = P F_1$
:$d_2 = P F_2$
Then $d$ is equal to the major axis of $K$.

Proof:
:400px
By the equidistance property of $K$:
:$d_1 + d_2 = d$
applies to all points $P$ on $K$.
Thus it also applies to the two vertices $V_1$ and $V_2$:
:$V_1 F_1 + V_1 F_2 = d$
:$V_2 F_1 + V_2 F_2 = d$
Adding:
:$V_1 F_1 + V_2 F_1 + V_1 F_2 + V_2 F_2 = 2 d$
But:
:$V_1 F_1 + V_2 F_1 = V_1 V_2$
:$V_1 F_2 + V_2 F_2 = V_1 V_2$
and so:
:$2 V_1 V_2 = 2 d$
By definition, the major axis is $V_1 V_2$.
Hence the result.
{{qed}}
Theorem: Equidistance of Hyperbola equals Transverse Axis

Let $K$ be an hyperbola whose foci are $F_1$ and $F_2$.
Let $P$ be an arbitrary point on $K$.
Let $d$ be the constant distance such that:
:$\size {d_1 - d_2} = d$
where:
:$d_1 = P F_1$
:$d_2 = P F_2$
Then $d$ is equal to the transverse axis of $K$.

Proof:
&nbsp;
:400px
By the equidistance property of $K$:
:$\size {d_1 - d_2} = d$
applies to all points $P$ on $K$.
Thus it also applies to the two vertices $V_1$ and $V_2$.
Observing the signs of $\size {d_1 - d_2}$ as appropriate:
:$V_1 F_2 - V_1 F_1 = d$
:$V_2 F_1 - V_2 F_2 = d$
Adding:
:$\paren {V_1 F_2 - V_2 F_2} + \paren {V_2 F_1 - V_1 F_1} = 2 d$
But:
:$V_1 F_2 - V_2 F_2 = V_1 V_2$
:$V_2 F_1 - V_1 F_1 = V_1 V_2$
and so:
:$2 V_1 V_2 = 2 d$
By definition, the transverse axis of $K$ is $V_1 V_2$.
Hence the result.
{{qed}}
Category:Hyperbolas
av6yuxgy28lhuu9wmw63l627zghzbcl
Theorem: Equivalence of Definitions of Ellipse

The following definitions of an ellipse are equivalent:

Proof:
Let $K$ be an ellipse aligned in a cartesian plane in reduced form.
Thus its foci are at $\tuple {\mathop \pm c, 0}$.
Let:
:the major axis of $K$ have length $2 a$
:the minor axis of $K$ have length $2 b$.
From Equation of Ellipse in Reduced Form, the equation of $K$ is:
:$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$
This has been derived from the equidistance property of $K$.
By definition, $c$ is the linear eccentricity of $K$.
From Linear Eccentricity of Ellipse from Major and Minor Axis:
:$a^2 - c^2 = b^2$
and so $K$ can be expressed as:
:$(1): \quad \dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$
It remains to be shown that $K$ possesses the focus-directrix property.
:400px
Let $D$ be the straight line whose equation is $x = \dfrac {a^2} c$.
It will be shown that the locus of points $P = \tuple {x, y}$ such that:
:the distance from $P$ to $F_1$ is $\dfrac c a$ of the distance from $P$ to $D$
is precisely equation $(1)$.
We have that:
:$P F_2 = \epsilon \paren {d - x}$
where:
:$\epsilon = \dfrac c a$
:$d = \dfrac {a^2} c$
Thus:
{{begin-eqn}}
{{eqn | l = \sqrt {\paren {x - c}^2 + y^2}
      | r = \frac c a \paren {\frac {a^2} c - x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {x - c}^2 + y^2
      | r = \paren {a - \frac c a x}^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x^2 - 2 c x + c^2 + y^2
      | r = a^2 - 2 c x + \frac {c^2} {a^2} x^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x^2 \paren {1 - \frac {c^2} {a^2} } + y^2
      | r = a^2 - c^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {a^2 - c^2} \frac {x^2} {a^2} + y^2
      | r = a^2 - c^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {x^2} {a^2} + \frac {y^2} {a^2 - c^2}
      | r = 1
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Ellipses
s9brzpii5h2fsb1ucj9cv7oap59swke
Theorem: Focus of Hyperbola from Transverse and Conjugate Axis

Let $K$ be a hyperbola whose transverse axis is $2 a$ and whose conjugate axis is $2 b$.
Let $c$ be the distance of the foci of $K$ from the center.
Then:
:$c^2 = a^2 + b^2$

Proof:
&nbsp;
:500px
Let the foci of $K$ be $F_1$ and $F_2$.
Let the vertices of $K$ be $V_1$ and $V_2$.
Let the covertices of $K$ be $C_1$ and $C_2$.
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 hyperbola.
This is true for all points on $K$.
In particular, it holds true for $V_2$, for example.
Thus:
{{begin-eqn}}
{{eqn | l = d
      | r = F_1 V_2 - F_2 V_2
      | c = 
}}
{{eqn | r = \paren {c + a} - \paren {c - a}
      | c = 
}}
{{eqn | r = 2 a
      | c = 
}}
{{end-eqn}}
{{finish|Some weird magic happens, and then:}}
{{begin-eqn}}
{{eqn | l = c
      | r = \sqrt {a^2 + b^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | r = c^2
      | l = a^2 + b^2
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Hyperbolas
nm1sq1wwb3bb0zztyacf19dyo773bs5
Theorem: Equation of Hyperbola in Reduced Form/Cartesian Frame/Parametric Form 1

The right-hand branch of $K$ can be expressed in parametric form as:
:$\begin {cases} x = a \cosh \theta \\ y = b \sinh \theta \end {cases}$

Proof:
Let the point $\tuple {x, y}$ satisfy the equations:
{{begin-eqn}}
{{eqn | l = x
      | r = a \cosh \theta
}}
{{eqn | l = y
      | r = b \sinh \theta
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \frac {x^2} {a^2} - \frac {y^2} {b^2}
      | r = \frac {\paren {a \cosh \theta}^2} {a^2} - \frac {\paren {b \sinh \theta}^2} {b^2}
      | c = 
}}
{{eqn | r =  \frac {a^2} {a^2} \cosh^2 \theta - \frac {b^2} {b^2} \sinh^2 \theta
      | c = 
}}
{{eqn | r = \cosh^2 \theta - \sinh^2 \theta
      | c = 
}}
{{eqn | r = 1
      | c = Difference of Squares of Hyperbolic Cosine and Sine
}}
{{end-eqn}}
{{qed}}
Theorem: Set of Condensation Points is Subset of Derivative

Let $T = \left({S, \tau}\right)$ be a topological space.
Let $A$ be a subset of $S$.
Then:
:$A^0 \subseteq A'$
where
:$A^0$ denotes the set of condensation points of $A$
:$A'$ denotes the derivative of $A$

Proof:
Let $x \in A^0$.
By definition of set of condensation points:
:$x$ is condensation point of $A$
By definition of condensation point:
:$x$ is limit point of $A$
By definition of derived set:
:$x \in A'$
{{qed}}
Theorem: Closure of Set of Condensation Points equals Itself

Let $T = \struct {S, \tau}$ be a topological space.
Let $A$ be a subset of $S$.
Then:
:$\paren {A^0}^- = A^0$
where
:$A^0$ denotes the set of condensation points of $A$
:$A^-$ denotes the closure of $A$

Proof:
By Set is Subset of its Topological Closure:
:$A^0 \subseteq \paren {A^0}^-$
To prove the equality by definition of set equality it suffices to show the inclusion:
:$\paren {A^0}^- \subseteq A^0$
Let $x \in \paren {A^0}^-$.
We will prove that
:$(1): \quad \forall U \in \tau: x \in U \implies A \cap U$ is uncountable
Let $U$ be an open subset of $S$ such that
:$x \in U$
By Condition for Point being in Closure:
:$A^0 \cap U \ne \O$
By definition of empty set:
:$\exists y: y \in A^0 \cap U$
By definition of intersection:
:$y \in A^0 \land y \in U$
By definition of set of condensation points:
:$y$ is condensation point of $A$
Then by definition of condensation point:
:$A \cap U$ is uncountable
We will prove that
:$x$ is limit point of $A$
Let $U$ be an open subset of $S$ such that
:$x \in U$
By $(1)$:
:$A \cap U$ is uncountable
:$A \cap \paren {U \setminus \set x}$ is uncountable
Thus:
:$A \cap \paren {U \setminus \set x} \ne \O$
Then by definition and $(1)$:
:$x$ is condensation point of $A$
Thus by definition of set of condensation points:
:$x \in A^0$
{{qed}}