hcju/proofwikiqa huggingface dataset:
=== 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}} === $S$ Subset Union of $T$ === Let $x \in S$. By definition of $T$: :$\set x \in T$ By Set is Subset of Union/Set of Sets: :$\ds \set x \subseteq \bigcup T$ By definition of singleton: :$x \in \set x$ Thus by definition of subset: :$\ds x \in \bigcup T$ {{qed|lemma}} Thus by definition of set equality: :$\ds \bigcup T = S$ {{qed}}
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: :$\displaystyle \int \frac {\d y} y = -\int \frac {x \rd x} {a^2 - x^2}$ 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}}
By Closure of Derivative is Derivative in T1 Space: :$\forall A \subseteq T: \left({A'}\right)^- = A'$ where :$A'$ denotes the derivative of $A$ :$\left({A'}\right)^-$ 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}}
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}}
The tangent line at $\tuple {Q, \map f \Q}$ on $f$ is defined as: :$\map {y_1} x = \paren {2 A Q + B} x + b_1$ where $2 A Q + B$ is the slope of the tangent line on the point $\tuple {Q, \map g Q}$ on $f$.    Substitute in the coordinates of the point $\tuple {Q, \map g Q}$ to $y_1$ and solve for $b_1$. This will reveal the $y$-intercept of $y_1$: {{begin-eqn}} {{eqn| l = A Q^2 + B Q + C_1      | r = \paren {2 A Q + B} Q + b_1      | c = The value of $y_1 = \map f Q$ }} {{eqn| l = A Q^2 + B Q + C_1 -2 A Q^2 - B Q      | r = b_1 }} {{eqn| l = -A Q^2 + C_1      | r = b_1 }} {{end-eqn}}      Continue by following the same steps for $y_2$ which is defined: :$\map {y_2} x = \paren {-2 A Q + B} x + b_2$ where $-2 A Q + B$ is the slope of the Tangent line at the point $\tuple {-Q, \map f {-Q} }$ on $f$. Substitute in the coordinates of the point $\paren {-Q, \map f {-Q} }$ to $y_2$. Use these values to solve for $b_2$, and this will reveal the $y$-intercept of $y_2$: {{begin-eqn}} {{eqn | l = A \paren {-Q}^2 + B \paren {-Q} + C_1       | r = \paren {-2 A Q + B} \paren {-Q} + b_2       | c = the value of $y_2 = \map f {-Q}$ }} {{eqn | l = A Q^2 - B Q + C_1 -2 A Q^2 + B Q       | r = b_2 }} {{eqn | l = -A Q^2 + C_1       | r = b_2 }} {{end-eqn}} The $y$-intercepts of both $y_1$ and $y_2$ have been shown to be equivalent. {{qed|lemma}} Since $b_1 = b_2$ redefine this value as $b$. The distance between $b$ and $C_1$ is $\size {C_1 - b}$. Let $\map g x = -A x^2 + B x + C_2$. Then the Tangent line at the point $\tuple {Q, \map g Q}$ on $g$ is defined as: :$\map {y_3} x = \paren {-2 A Q + B} x + b_3$ where $-2 A Q + B$ is the slope of the tangent line at $\tuple {Q, \map g Q}$ on $g$. Solve for $b_3$ using the same methods used for $y_1$ and $y_2$. This will reveal the $y$-intercept of $y_3$: :$b_3 = A Q^2 + C_2$ The result also follows for the Tangent line $\tuple {-Q, \map g {-Q} }$ on $g$ which is defined: :$y_4 = \paren {-2 A Q + B} x + b_4$ Solving for $b_4$ yields the result: :$b_4 = A Q^2 + C_2$ The $y$-intercepts of both $y_3$ and $y_4$ have been shown to be equivalent. {{qed|lemma}} Notice that the derivatives of $f$ and $g$ satisfy: {{begin-eqn}} {{eqn | l = \map {g'} Q       | r = \map {f'} {-Q} }} {{eqn | l = \map {g'} {-Q}       | r = \map {f'} Q }} {{end-eqn}} Then it must be true that: :$y_1 = y_4$ and $y_2 = y_3$ and the functions $y_1$, $y_2$, $y_3$, and $y_4$ share the same $y$-intercept. {{qed|lemma}} Redefine this the $y$-intercepts of the tangent lines as $b$. Solve for $C_2$ to determine the vertical translation of $\map g x$: {{begin-eqn}} {{eqn | l = C_2       | r = \paren {-A Q^2 + C_1} - A Q^2 }} {{eqn | l = C_2       | r = -2 A Q^2 + C_1 }} {{end-eqn}} Therefore the function: :$\map g x = -A x^2 + B x - \paren {2 A Q^2 + C_1}$ will have tangent lines equivalent to the tangent lines on $\map f x$ at the points $\tuple {Q, \map f Q}$, and $\tuple {-Q, \map f {-Q} }$. {{qed}} Category:Analytic Geometry ao18o8hiwqw6up7edqzndcsegp1i1r5
=== Sufficient Condition === Let $T$ be a $T_0$ space. Let $x, y \in S$ such that :$x \ne y$ Aiming for a contradiction suppose that :$x \in \left\{{y}\right\}^- \land y \in \left\{{x}\right\}^-$ Then: :$\left\{{x}\right\} \subseteq \left\{{y}\right\}^- \land \left\{{y}\right\} \subseteq \left\{{x}\right\}^-$ By Topological Closure of Subset is Subset of Topological Closure: :$\left\{{x}\right\}^- \subseteq \left({\left\{{y}\right\}^-}\right)^- \land \left\{{y}\right\}^- \subseteq \left({\left\{{x}\right\}^-}\right)^-$ By Closure of Topological Closure equals Closure: :$\left\{{x}\right\}^- \subseteq \left\{{y}\right\}^- \land \left\{{y}\right\}^- \subseteq \left\{{x}\right\}^-$ Then by definition of set equality: :$\left\{{x}\right\}^- = \left\{{y}\right\}^-$ By Characterization of $T_0$ Space by Distinct Closures of Singletons: :$\left\{{x}\right\}^- \ne \left\{{y}\right\}^-$ This contradicts the equality. Thus the result by Proof by Contradiction {{qed|lemma}} === Necessary Condition === Assume that: :$(1): \quad \forall x, y \in S: x \ne y \implies x \notin \left\{{y}\right\}^- \lor y \notin \left\{{x}\right\}^-$ By Characterization of $T_0$ Space by Distinct Closures of Singletons it suffices to prove :$\forall x, y \in S: x \ne y \implies \left\{{y}\right\}^- \ne \left\{{x}\right\}^-$ Let $x, y \in S$ such that :$x \ne y$ Aiming for a contradiction suppose that :$(2): \quad \left\{{y}\right\}^- = \left\{{x}\right\}^-$ By definition of singleton: :$x \in \left\{{x}\right\} \land y \in \left\{{y}\right\}$ By Set is Subset of its Topological Closure :$\left\{{x}\right\} \subseteq \left\{{x}\right\}^- \land \left\{{y}\right\} \subseteq \left\{{y}\right\}^-$ Then by definition of subset and $(2)$: :$x \in \left\{{y}\right\}^- \land y \in \left\{{x}\right\}^-$ This contradicts the assumption $(1)$. Thus the result by Proof by Contradiction. {{qed}}
=== 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: :$\left({\exists U \in \tau: x \in U \land y \notin U}\right) \lor \left({\exists U \in \tau: x \notin U \land y \in U}\right)$ WLOG: Suppose :$\exists U \in \tau: x \in U \land y \notin U$ By definition of singleton: :$x \in \left\{{x}\right\}$ By Set is Subset of its Topological Closure: :$\left\{{x}\right\} \subseteq \left\{{x}\right\}^-$ Then by definition of subset: :$x \in \left\{{x}\right\}^-$ Hence by definition of intersection: :$\left\{{x}\right\}^- \cap U \ne \varnothing$ By definition of singleton: :$\forall z: z \in \left\{{y}\right\} \implies z = y$ Then by definition of intersection: :$\left\{{y}\right\} \cap U = \varnothing$ Hence by Disjoint Open Sets remain Disjoint with one Closure :$\left\{{y}\right\}^- \cap U = \varnothing$ Thus: :$\left\{{x}\right\}^- \ne \left\{{y}\right\}^-$ {{qed|lemma}} === Necessary Condition === Assume that: :$(1): \quad \forall x, y \in S: x \ne y \implies \left\{{x}\right\}^- \ne \left\{{y}\right\}^-$ By Characterization of $T_0$ Space by Closed Sets it suffices to prove that :$\left({\exists F \subseteq S: F}\right.$ is closed $\left. {\land x \in F \land y \notin F}\right)$ or :$\left({\exists F \subseteq S: F}\right.$ is closed $\left. {\land x \notin F \land y \in F}\right)$ Then assume: :$(2): \quad \forall F \subseteq S: F$ is closed $\land x \in F \implies y \in F$ Define $F := \left\{{y}\right\}^-$ By Topological Closure is Closed: :$F$ is closed. We will prove now that :$x \notin F$ Aiming for a contradiction suppose that :$x \in F$ Then: :$\left\{{x}\right\} \subseteq F$ By Topological Closure of Subset is Subset of Topological Closure: :$\left\{{x}\right\}^- \subseteq F^-$ By Set is Closed iff Equals Topological Closure: :$(3): \quad \left\{{x}\right\}^- \subseteq F$ By definition of singleton: :$x \in \left\{{x}\right\}$ By Set is Subset of its Topological Closure: :$\left\{{x}\right\} \subseteq \left\{{x}\right\}^-$ Then by definition of subset: :$x \in \left\{{x}\right\}^-$ Because $\left\{{x}\right\}^-$ is closed therefore by $(2)$: :$y \in \left\{{x}\right\}^-$ Then: :$\left\{{y}\right\} \subseteq \left\{{x}\right\}^-$ By Topological Closure of Subset is Subset of Topological Closure: :$F \subseteq \left({\left\{{x}\right\}^-}\right)^-$ Then by Closure of Topological Closure equals Closure: :$F \subseteq \left\{{x}\right\}^-$ Hence by $(3)$ and definition of set equality: :$F = \left\{{x}\right\}^-$ This contradicts the assumption $(1)$. Thus $x \notin F$. By definition of singleton: :$y \in \left\{{y}\right\}$ By Set is Subset of its Topological Closure: :$\left\{{y}\right\} \subseteq \left\{{y}\right\}^- = F$ Thus by definition of subset: :$y \in F$ Hence :$\exists F \subseteq S: F$ is closed $\land x \notin F \land y \in F$ {{qed}}
=== 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: :$\left({\exists U \in \tau: x \in U \land y \notin U}\right) \lor \left({\exists U \in \tau: x \notin U \land y \in U}\right)$ WLOG: Suppose: :$\exists U \in \tau: x \in U \land y \notin U$ By definition: :$\complement_S\left({U}\right)$ is closed where $\complement_S\left({U}\right)$ denotes the relative complement of $U$ in $S$. By definition of relative complement: :$x \notin \complement_S\left({U}\right) \land y \in \complement_S\left({U}\right)$ Thus: :$\exists F \subseteq S: F$ is closed $\land\, x \notin F \land y \in F$ {{qed|lemma}} === Necessary Condition === This statement follows mutatis mutandis. {{qed}}
: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 $P = \left({x, y}\right)$ 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}}
: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}}
&nbsp; :400px By the equidistance property of $K$: :$\left\lvert{d_1 - d_2}\right\rvert = d$ applies to all points $P$ on $K$. Thus it also applies to the two vertices $V_1$ and $V_2$. Observing the signs of $\left\lvert{d_1 - d_2}\right\rvert$ as appropriate: :$V_1 F_2 - V_1 F_1 = d$ :$V_2 F_1 - V_2 F_2 = d$ Adding: :$\left({V_1 F_2 - V_2 F_2}\right) + \left({V_2 F_1 - V_1 F_1}\right) = 2 d$ 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 slpd59kn99njqsfae0azz17ox2rwikm
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 $K$. From Focus 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 5qzqtap264tdqssj29kxqz3bdgmv01k
Let $K$ be an hyperbola aligned in a cartesian plane in reduced form. Thus its foci are at $\tuple {\mathop \pm c, 0}$. Let: :the transverse axis of $K$ have length $2 a$ :the conjugate axis of $K$ have length $2 b$. From Equation of Hyperbola 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 $K$. From Focus of Hyperbola from Transverse and Conjugate Axis: :$c^2 - a^2 = b^2$ and so $K$ can be expressed as: :$(1): \quad \dfrac {x^2} {a^2} - \dfrac {y^2} {c^2 - a^2} = 1$ 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: :$P F_2 = \dfrac c a d$ where: :$d = x - \dfrac {a^2} c$ Thus: {{begin-eqn}} {{eqn | l = \sqrt {\paren {c - a}^2 + y^2}       | r = \frac c a \paren {x - \frac {a^2} c}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {c - x}^2 + y^2       | r = \paren {\frac c a x - a}^2       | c =  }} {{eqn | ll= \leadsto       | l = c^2 - 2 c x + x^2 + y^2       | r = \frac {c^2} {a^2} x^2 - 2 c x + a^2       | c =  }} {{eqn | ll= \leadsto       | l = x^2 \paren {\frac {c^2} {a^2} - 1} - y^2       | r = c^2 - a^2       | c =  }} {{eqn | ll= \leadsto       | l = \paren {c^2 - a^2} \frac {x^2} {a^2} - y^2       | r = c^2 - a^2       | c =  }} {{eqn | ll= \leadsto       | l = \frac {x^2} {a^2} - \frac {y^2} {c^2 - a^2}       | r = 1       | c =  }} {{end-eqn}} {{qed}} Category:Hyperbolas qzjk6m9b2rbmlbm57xs19cx319oimw0
&nbsp; :500px 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: :$c^2 a^2 = b^2$ {{WLOG}}, let us choose a point $P$ such that $F_1 P > F_2 P$. Then: {{begin-eqn}} {{eqn | l = \sqrt {\paren {x + c}^2 + y^2} - \sqrt {\paren {x - c}^2 + y^2}       | r = d = 2 a       | c = Pythagoras's Theorem }} {{eqn | ll= \leadsto       | l = \sqrt {\paren {x + c}^2 + y^2}       | r = 2 a + \sqrt {\paren {x - c}^2 + y^2}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x + c}^2 + y^2       | r = \paren {2 a + \sqrt {\paren {x - c}^2 + y^2} }^2       | c = squaring both sides }} {{eqn | ll= \leadsto       | l = x^2 + 2 c x + c^2 + y^2       | r = 4 a^2 + 4 a \sqrt {\paren {x - c}^2 + y^2} + \paren {x - c}^2 + y^2       | c = expanding }} {{eqn | ll= \leadsto       | l = x^2 + 2 c x + c^2 + y^2       | r = 4 a^2 + 4 a \sqrt {\paren {x - c}^2 + y^2} + x^2 - 2 c x + c^2 + y^2       | c = further expanding }} {{eqn | ll= \leadsto       | l = c x - a^2       | r = a \sqrt {\paren {x - c}^2 + y^2}       | c = gathering terms and simplifying }} {{eqn | ll= \leadsto       | l = \paren {c x - a^2}^2       | r = a^2 \paren {\paren {x - c}^2 + y^2}^2       | c = squaring both sides }} {{eqn | ll= \leadsto       | l = c^2 x^2 - 2 c x a^2 + a^4       | r = a^2 x^2 - 2 c x a^2 + a^2 c^2 + a^2 y^2       | c = expanding }} {{eqn | ll= \leadsto       | l = c^2 x^2 + a^4       | r = a^2 x^2 + a^2 c^2 + a^2 y^2       | c = simplifying }} {{eqn | ll= \leadsto       | l = a^2 c^2 - a^4       | r = c^2 x^2 - a^2 x^2 - a^2 y^2       | c = gathering terms }} {{eqn | ll= \leadsto       | l = a^2 \paren {c^2 - a^2}       | r = \paren {c^2 - a^2} x^2 - a^2 y^2       | c = simplifying }} {{eqn | ll= \leadsto       | l = a^2 b^2       | r = b^2 x^2 - a^2 y^2       | c = substituting $c^2 - a^2 = b^2$ from $(2)$ }} {{eqn | ll= \leadsto       | l = 1       | r = \frac {x^2} {a^2} - \frac {y^2} {b^2}       | c = dividing by $a^2 b^2$ }} {{end-eqn}} {{qed}}
&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 5ciwgbtn56ylvhku9v7h5tt00apkaoi
Let the point $\tuple {x, y}$ satisfy the equations: :$x = a \cosh \theta$ :$y = b \sinh \theta$ 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}}
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}}