avewright/proofwiki-math huggingface dataset:
Theorem: Orthocenter of Self-Conjugate Triangle

Let $\CC$ be a circle.
Let $\triangle PQR$ be a '''self-conjugate triangle with respect to $\CC$'''.
Then the orthocenter of $\triangle PQR$ is the center of $\CC$.

Proof:
By definition of self-conjugate triangle:
:$PR$ is the polar of $Q$
:$QR$ is the polar of $P$
and from Self-Conjugate Triangle needs Two Sides to be Specified:
:$PQ$ is the polar of $R$
all with respect to $\CC$.
Let $O$ be the center of $\CC$.
Then from Polar of Point is Perpendicular to Line through Center:
:$PQ \perp OR$
:$QR \perp OP$
:$PR \perp OQ$
That is: $OP$, $OQ$ and $OR$ are lines perpendicular to the sides of $\triangle PQR$ which all pass through $O$.
Hence, by definition, $O$ is the orthocenter of $\triangle PQR$.
{{Qed}}
Theorem: Hat-Check Distribution Gives Rise to Probability Mass Function

Let $X$ be a discrete random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $X$ have the hat-check distribution with parameter $n$ (where $n > 0$).
Then $X$ gives rise to a probability mass function.

Proof:
By definition:
:$\Img X = \set {0, 1, \ldots, n}$
:$\ds \map \Pr {X = k} = \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}$
Then:
{{begin-eqn}}
{{eqn | l = \map \Pr \Omega
      | r = \sum_{k \mathop = 0}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^n \dfrac 1 {\paren {n - k }! } \dfrac {n! k!} {n! k!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}
      | c = multiplying by $1$
}}
{{eqn | r = \sum_{k \mathop = 0}^n \dbinom n k \dfrac {k!} {n!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}
      | c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \dfrac 1 {n!} \sum_{k \mathop = 0}^n \dbinom n k k! \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}
      | c = 
}}
{{eqn | r = \dfrac 1 {n!} n!
      | c = Sum over k of r Choose k by -1^r-k by Polynomial
}}
{{eqn | r = 1
      | c = 
}}
{{end-eqn}}
So $X$ satisfies $\map \Pr \Omega = 1$, and hence the result.
{{Qed}}
Category:Hat-Check Distribution
Category:Hat-Check Problem
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Theorem: Expectation of Hat-Check Distribution

Let $X$ be a discrete random variable with the Hat-Check distribution with parameter $n$.
Then the expectation of $X$ is given by:
:$\expect X = n - 1$

Proof:
From the definition of expectation:
:$\ds \expect X = \sum_{x \mathop \in \Img X} x \map \Pr {X = x}$
By definition of hat-check distribution:
:$\ds \expect X = \sum_{k \mathop = 0}^n k \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}$
Then:
{{begin-eqn}}
{{eqn | l = \expect X
      | r = \sum_{k \mathop = 1}^n k \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}
      | c = as the $k = 0$ term vanishes
}}
{{eqn | r = \sum_{y \mathop = n - 1}^0 \paren {n - y } \dfrac 1 {y!} \sum_{s \mathop = 0}^{n - y} \dfrac {\paren {-1}^s} {s!}
      | c = Let $y = n - k$
}}
{{eqn | r = n \sum_{y \mathop = 0}^{n - 1} \dfrac 1 {y!} \sum_{s \mathop = 0}^{n - y} \dfrac {\paren {-1}^s} {s!} - \sum_{y \mathop = 0}^{n - 1} \dfrac y {y!} \sum_{s \mathop = 0}^{n - y} \dfrac {\paren {-1}^s} {s!}
      | c = 
}}
{{eqn | r = n \sum_{k \mathop = 1}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - \sum_{k \mathop = 1}^n \dfrac {n - k} {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}
      | c = Let $y = n - k$
}}
{{eqn | r = n \sum_{k \mathop = 1}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} + \dfrac n {n!} - \dfrac n {n!} - \sum_{k \mathop = 1}^n \dfrac {n - k} {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}
      | c = adding $0$
}}
{{eqn | r = n \sum_{k \mathop = 0}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - \sum_{k \mathop = 0}^n \dfrac {n - k} {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}
      | c = 
}}
{{eqn | r = n \sum_{k \mathop = 0}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - \sum_{k \mathop = 0}^{n - 1} \dfrac {n - k} {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}
      | c = as the $k = n$ term vanishes
}}
{{eqn | r = n \sum_{k \mathop = 0}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - \sum_{k \mathop = 0}^{n - 1} \dfrac 1 {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}
      | c = 
}}
{{eqn | r = n - 1
      | c = Hat-Check Distribution Gives Rise to Probability Mass Function
}}
{{end-eqn}}
{{qed}}
Theorem: Jump Rule

Let $f : \R \to \R$ be a piecewise continuously differentiable real function with a discontinuity at $c \in \R$.
Suppose the limits $\map f {c^+}, \map f {c^-}, \map {f'} {c^+}, \map {f'} {c^-}$ exist.
Let $T \in \map {\DD'} \R$ be a Schwartz distribution associated with $f$.
Then in the distributional sense we have that:
:$T_f' = T_{f'} + \paren {\map f {c^+} - \map f {c^-}} \delta_c$
where $\delta_c$ is the Dirac delta distribution.

Proof:
Let $\phi \in \map \DD {\R}$ be a test function with its support on $\closedint a b \subset \R$.
Let $c \in \closedint a b$.
Then:
{{begin-eqn}}
{{eqn | l = \map {T_f'} \phi
      | r = -\map {T_f} {\phi'}
      | c = {{Defof|Distributional Derivative}}
}}
{{eqn | r = -\int_a^b \map f x \map {\phi'} x \rd x
}}
{{eqn | r =  \int_a^c \map f x \map {\phi'} x \rd x - \int_c^b \map f x \map {\phi'} x \rd x
}}
{{eqn | r = \map f a \map \phi a - \map f {c^-} \map \phi {c^-} + \int_a^c \map {f'} x \map \phi x + \map f {c^+} \map \phi {c^+} - \map f b \map \phi b + \int_c^b \map {f'} x \map \phi x \rd x
      | c = Integration by parts
}}
{{eqn | r = \int_a^b \map {f'} x \map \phi x \rd x + \paren {\map f {c^+} - \map f {c^-} } \map \phi c
      | c = {{Defof|Test Function}}
}}
{{eqn | r = \map {T_{f'} } \phi + \paren {\map f {c^+} - \map f {c^-} } \map {\delta_c} \phi
      | c = {{Defof|Dirac Delta Distribution}}
}}
{{eqn | r = \map {\paren {T_{f'} + \paren {\map f {c^+} - \map f {c^-} }\delta_c} } \phi
      | c = Schwartz distribution is linear mapping
}}
{{end-eqn}}
{{qed}}
{{Research|It this strict enough? We do not have a limit in the beginning, but later we have to introduce it. Is this the aspect of the distribution or of the Riemannian integral.}}
Theorem: Characteristic of Interior Point of Circle whose Center is Origin

Let $\CC$ be a circle of radius $r$ whose center is at the origin $O$ of a Cartesian plane.
Let $P = \tuple {x, y}$ be a point in the plane of $\CC$.
Then $P$ is in the interior of $\CC$ {{iff}}:
:$x^2 + y^2 - r^2 < 0$

Proof:
Let $d$ be the distance of $P$ from $O$.
{{begin-eqn}}
{{eqn | l = d
      | r = \sqrt {\paren {x - 0}^2 + \paren {y - 0}^2}
      | c = Distance Formula
}}
{{eqn | ll= \leadsto
      | l = d^2
      | r = x^2 + y^2
      | c = 
}}
{{end-eqn}}
Then by definition of interior of $\CC$:
:$P$ is in the interior of $\CC$ {{iff}} $d^2 < r^2$
and the result follows.
{{qed}}
Theorem: Length of Tangent from Point to Circle center Origin

Let $\CC$ be a circle of radius $r$ whose center is at the origin $O$ of a Cartesian plane.
Let $P = \tuple {x, y}$ be a point in the plane of $\CC$ in the exterior of $\CC$.
Let $PT$ be a tangent to $\CC$ from $P$ such that $T$ is the point of tangency.
Then the length of $PT$ is given by:
:$PT^2 = x^2 + y^2 - r^2$

Proof:
Let $\NN$ be the normal to $\CC$ at the point $T$.
From Normal to Circle passes through Center, $\NN$ passes through $O$.
By definition of the normal to $\CC$, $\NN$ is perpendicular to $PT$.
Hence $OT$, $PT$ and $OP$ form a right triangle whose hypotenuse is $OP$.
As $OT$ is a line segment coinciding with a radius of $\CC$:
:$OT = r$
The length of $OP$ comes from the Distance Formula:
:$OP = \sqrt {\paren {x - 0}^2 + \paren {y - 0}^2} = \sqrt {x^2 + y^2}$
Hence we have:
{{begin-eqn}}
{{eqn | l = OP^2
      | r = OT^2 + PT^2
      | c = Pythagoras's Theorem
}}
{{eqn | ll= \leadsto
      | l = PT^2
      | r = x^2 + y^2 - r^2
      | c = substituting for $OP$ and $OT$ from above
}}
{{end-eqn}}
{{qed}}
Theorem: Equation of Tangents to Circle from Point

Let $\CC$ be a circle embedded in the Cartesian plane of radius $r$ with its center located at the origin.
Let $P = \tuple {x_0, y_0}$ be a point in the plane of $\CC$ which is outside $\CC$.
The tangents to $\CC$ which pass through $P$ can be described using the equation:
:$\paren {x x_0 + y y_0 - r^2}^2 = \paren {x^2 + y^2 - r^2} \paren { {x_0}^2 + {y_0}^2 - r^2}$

Proof:
From Equation of Circle center Origin, $\CC$ can be described as:
:$x^2 + y^2 = r^2$
Let $\LL$ be an arbitrary straight line through $P$ which intersects $\CC$ at $U$ and $V$.
Let $Q = \tuple{x, y}$ be an arbitrary point on $\LL$.
Let $k$ be the position-ratio of one of the points $U$ and $V$ with respect to $P$ and $Q$.
By Joachimsthal's equation:
:$(1): \quad k^2 \paren {x^2 + y^2 - r^2} + 2 k \paren {x x_0 + y y_0 - r^2} + \paren { {x_0}^2 + {y_0}^2 - r^2} = 0$
which is a quadratic in $k$.
When $\LL$ is tangent to $\CC$, the points $U$ and $V$ coincide.
Hence $(1)$ has equal roots.
From Solution to Quadratic Equation, it follows that the discriminant of $(1)$ is zero.
That is:
:$\paren {x x_0 + y y_0 - r^2}^2 - \paren {x^2 + y^2 - r^2} \paren { {x_0}^2 + {y_0}^2 - r^2} = 0$
{{qed}}
Theorem: Condition of Tangency to Circle whose Center is Origin

Let $\CC$ be a circle embedded in the Cartesian plane of radius $r$ with its center located at the origin.
Let $\LL$ be a straight line in the plane of $\CC$ whose equation is given by:
:$(1): \quad l x + m y + n = 0$
such that $l \ne 0$.
Then $\LL$ is tangent to $\CC$ {{iff}}:
:$\paren {l^2 + m^2} r^2 = n^2$

Proof:
From Equation of Circle center Origin, $\CC$ can be described as:
:$(2): \quad x^2 + y^2 = r^2$
Let $\LL$ intersect with $\CC$.
To find where this happens, we find $x$ and $y$ which satisfy both $(1)$ and $(2)$.
So:
{{begin-eqn}}
{{eqn | n = 1
      | l = l x + m y + n
      | r = 0
      | c = Equation for $\LL$
}}
{{eqn | ll= \leadsto
      | l = x
      | r = -\dfrac {m y} l - \dfrac n l
      | c = rearranging
}}
{{eqn | ll= \leadsto
      | l = \paren {-\dfrac {m y} l - \dfrac n l}^2 + y^2
      | r = r^2
      | c = substituting for $x$ in $(2)$
}}
{{eqn | ll= \leadsto
      | l = \paren {-m y - n}^2 + l^2 y^2
      | r = l^2 r^2
      | c = multiplying by $l^2$
}}
{{eqn | ll= \leadsto
      | l = m^2 y^2 + 2 m n y + n^2 + l^2 y^2
      | r = l^2 r^2
      | c = multiplying out
}}
{{eqn | n = 3
      | ll= \leadsto
      | l = \paren {l^2 + m^2} y^2 + 2 m n y + \paren {n^2 - l^2 r^2}
      | r = 0
      | c = rearranging
}}
{{end-eqn}}
This is a quadratic in $y$.
This corresponds to the two points of intersection of $\LL$ with $\CC$.
When $\LL$ is tangent to $\CC$, these two points coincide.
Hence $(3)$ has equal roots.
From Solution to Quadratic Equation, this happens when the discriminant of $(3)$ is zero.
That is:
{{begin-eqn}}
{{eqn | l = m^2 n^2
      | r = \paren {l^2 + m^2} \paren {n^2 - l^2 r^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = l^2 n^2 - l^2 m^2 r^2 - l^4 r^2
      | r = 0
      | c = multiplying out and simplifying
}}
{{eqn | ll= \leadsto
      | l = \paren {l^2 + m^2} r^2
      | r = n^2
      | c = as $l^2 \ne 0$
}}
{{end-eqn}}
{{qed}}
Theorem: Product of Smooth Function and Dirac Delta Distribution

Let $a \in \R^d$ be a vector in real Euclidean space.
Let $\delta_a \in \map {\DD'} {\R^d}$ be the Dirac delta distribution.
Let $\alpha \in \map {C^\infty} {\R^d}$ be a smooth real function.
Then in the Schwartz distributional sense we have that:
:$\alpha \delta_a = \map \alpha a \delta_a$

Proof:
Let $\phi \in \map \DD {\R^d}$ be a test function.
Then:
{{begin-eqn}}
{{eqn | l = \map {\alpha \delta_a} \phi
      | r = \map {\delta_a} {\alpha \phi}
      | c = {{Defof|Multiplication of Schwartz Distribution by Smooth Function}}
}}
{{eqn | r = \map {\paren {\alpha \phi} } a
      | c = {{Defof|Dirac Delta Distribution}}
}}
{{eqn | r = \map \alpha a \map \phi a
      | c = {{Defof|Pointwise Multiplication of Real-Valued Functions}}
}}
{{eqn | r = \map \alpha a \map {\delta_a} \phi
      | c = {{Defof|Dirac Delta Distribution}}
}}
{{end-eqn}}
{{qed}}
Theorem: Circle is Curve of Second Degree

The circle is a curve of degree $2$.

Proof:
From Equation of Circle in Cartesian Plane, a circle can be expressed in the form:
:$x^2 + y^2 - 2 a x - 2 b y + a^2 + b^2 - R^2 = 0$
where $\tuple {a, b}$ is the center and $R$ is the radius.
This is a quadratic equation in $2$ variables.
Hence the result by definition of degree of curve.
{{qed}}
{{explain|Sommerville is not rigorous about defining exactly what the degree of a curve is, so the definition needs to wait until such time as I can hunt it down from another source.}}
Theorem: Characteristic of Quadratic Equation that Represents Circle

A '''quadratic equation in $2$ variables''':
:$a x^2 + b y^2 + 2 h x y + 2 g x + 2 f y + c = 0$
describes a circle embedded in the Cartesian plane {{iff}}:
:$(1): \quad h = 0$
:$(2): \quad a = b \ne 0$

Proof:
=== Necessary Condition ===
Consider the equation of a circle:
:$A \paren {x^2 + y^2} + B x + C y + D = 0$
which is the equation of a circle with radius $R$ and center $\tuple {a, b}$, where:
:$R = \dfrac 1 {2 A} \sqrt {B^2 + C^2 - 4 A D}$
:$\tuple {a, b} = \tuple {\dfrac {-B} {2 A}, \dfrac {-C} {2 A} }$
provided:
:$A > 0$
:$B^2 + C^2 \ge 4 A D$
Thus it is seen that the coefficient for $x^2$ is the same as that for $y^2$, and there is no $x y$ term.
As can be seen, this matches the form as described: $a = b$ and $h = 0$.
{{qed|lemma}}
Theorem: Angle of Intersection of Circles equals Angle between Radii

Let $\CC$ and $\CC'$ be circles whose centers are at $C$ and $C'$ respectively.
Let $\CC$ and $\CC'$ intersect at $A$ and $B$.
The angle of intersection of $\CC$ and $\CC'$ is equal to the angle between the radii to the point of intersection.

Proof:
From Normal to Circle passes through Center, the straight line passing through the center of a circle is normal to that circle.
Hence the radii $CA$ and $C'A$ are perpendicular to the tangents to $\CC$ and $\CC'$ respectively.
:540px
Thus, with reference to the above diagram, we have that:
:$\angle FAC = \angle DAC'$
as both are right angles.
Hence:
{{begin-eqn}}
{{eqn | l = \angle FAC'
      | r = \angle FAC + \angle CAC'
      | c = 
}}
{{eqn | r = \angle FAD + \angle DAC'
      | c = from above
}}
{{eqn | ll= \leadsto
      | l = \angle FAC + \angle CAC'
      | r = \angle FAD + \angle FAC
      | c = as $\angle FAC = \angle DAC'$
}}
{{eqn | ll= \leadsto
      | l = \angle FAD
      | r = \angle CAC'
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Angle of Intersection of Circles is Equal at Both Points

Let $\CC$ and $\CC'$ be circles whose centers are at $C$ and $C'$ respectively.
Let $\CC$ and $\CC'$ intersect at $A$ and $B$.
The angle of intersection of $\CC$ and $\CC'$ at $A$ is equal to the angle of intersection of $\CC$ and $\CC'$ at $B$.

Proof:
Consider the two triangles $CAC'$ and $CBC'$.
:540px
By definition of radius:
:$CA = CB$ and $C'A = C'B$
{{begin-eqn}}
{{eqn | l = CA
      | r = CB
      | c = {{Defof|Radius of Circle}}
}}
{{eqn | l = C'A
      | r = C'B
      | c = {{Defof|Radius of Circle}}
}}
{{eqn | ll= \leadsto
      | l = \triangle CAC'
      | r = \triangle CBC'
      | c = Triangle Side-Side-Side Congruence
}}
{{eqn | ll= \leadsto
      | l = \angle CAC'
      | r = \angle CBC'
      | c = 
}}
{{end-eqn}}
The result follows from Angle of Intersection of Circles equals Angle between Radii.
{{qed}}
Theorem: Equation of Circle/Cartesian/Formulation 1

The equation of a circle embedded in the Cartesian plane with radius $R$ and center $\tuple {a, b}$ can be expressed as:
:$\paren {x - a}^2 + \paren {y - b}^2 = R^2$

Proof:
Let the point $\tuple {x, y}$ satisfy the equation:
:$(1): \quad \paren {x - a}^2 + \paren {y - b}^2 = R^2$
By the Distance Formula, the distance between this $\tuple {x, y}$ and $\tuple {a, b}$ is:
:$\sqrt {\paren {x - a}^2 + \paren {y - b}^2}$
But from equation $(1)$, this quantity equals $R$.
Therefore the distance between points satisfying the equation and the center is constant and equal to the radius.
Thus $\tuple {x, y}$ lies on the circumference of a circle with radius $R$ and center $\tuple {a, b}$.
Now suppose that $\tuple {x, y}$ does not satisfy the equation:
:$\paren {x - a}^2 + \paren {y - b}^2 = R^2$
Then by the same reasoning as above, the distance between $\tuple {x, y}$ and $\tuple {a, b}$ does not equal $R$.
Therefore $\tuple {x, y}$ does not lie on the circumference of a circle with radius $R$ and center $\tuple {a, b}$.
Hence it follows that the points satisfying $(1)$ are exactly those points which are the circle in question.
{{qed}}
Theorem: Condition for Circles to be Orthogonal

Let $\CC_1$ and $\CC_2$ be circles embedded in a Cartesian plane.
Let $\CC_1$ and $\CC_2$ be described by Equation of Circle in Cartesian Plane as:
{{begin-eqn}}
{{eqn | q = \CC_1
      | l = x^2 + y^2 + 2 \alpha_1 x + 2 \beta_1 y + c_1
      | r = 0
      | c = 
}}
{{eqn | q = \CC_2
      | l = x^2 + y^2 + 2 \alpha_2 x + 2 \beta_2 y + c_2
      | r = 0
      | c = 
}}
{{end-eqn}}
Then $\CC_1$ and $\CC_2$ are orthogonal {{iff}}:
:$2 \alpha_1 \alpha_2 + 2 \beta_1 \beta_2 = c_1 + c_2$

Proof:
When $\CC_1$ and $\CC_2$ are orthogonal, the distance between their centers forms the hypotenuse of a right triangle whose legs are equal to the radii.
From Equation of Circle in Cartesian Plane: Formulation 3, the radii $r_1$ and $r_2$ of $\CC_1$ and $\CC_2$ respectively are given by:
{{begin-eqn}}
{{eqn | l = c_1
      | r = \alpha_1^2 + \beta_1^2 - r_1^2
}}
{{eqn | l = c_2
      | r = \alpha_2^2 + \beta_2^2 - r_2^2
}}
{{end-eqn}}
Hence we have:
{{begin-eqn}}
{{eqn | l = \paren {\alpha_1 - \alpha_2}^2 + \paren {\beta_1 - \beta_2}^2
      | r = r_1^2 + r_2^2
      | c = Pythagoras's Theorem and Distance Formula
}}
{{eqn | ll= \leadsto
      | l = \alpha_1^2 + \alpha_2^2 - 2 \alpha_1 \alpha_2 + \beta_1^2 + \beta_2^2 - 2 \beta_1 \beta_2
      | r = r_1^2 + r_2^2
      | c = multiplying out
}}
{{eqn | ll= \leadsto
      | l = 2 \alpha_1 \alpha_2 + 2 \beta_1 \beta_2
      | r = \paren {\alpha_1^2 + \beta_1^2 - r_1^2} + \paren {\alpha_2^2 + \beta_2^2 - r_2^2}
      | c = multiplying out
}}
{{eqn | ll= \leadsto
      | l = 2 \alpha_1 \alpha_2 + 2 \beta_1 \beta_2
      | r = c_1 + c_2
      | c = from above
}}
{{end-eqn}}
{{qed}}
Theorem: Test Function with Vanishing Partial Derivative

Let $\phi \in \map \DD {\R^2}$ be a test function such that:
:$\tuple {x, y} \stackrel \phi {\longrightarrow} \map \phi {x, y}$
Suppose $\phi$ is a solution to the following partial differential equation:
:$\dfrac {\partial \phi} {\partial x} = 0$
Then $\phi$ is identically $0$.

Proof:
$\dfrac {\partial \phi} {\partial x} = 0$ implies that:
:$\forall x \in \R : \map \phi {x, y} = \map C y$
where $C : \R \to \C$ is a complex-valued function.
By definition, $\phi$ is a test function.
Hence, $\phi$ must have a compact support $\Omega \subset \R^2$.
Let $\map {B^-_\epsilon} 0 \subset \R^2$ be a closed ball in Euclidean space such that:
:$\Omega \subseteq \map {B^-_\epsilon} 0$
By Closed Ball in Euclidean Space is Compact, $\map {B^-_\epsilon} 0$ is compact.
Then $\map {B^-_\epsilon} 0$ also qualifies as a compact support of $\phi$.
By definition of a test function:
:$\forall y \in \R : \size y > \epsilon \implies \map \phi {x, y} = 0$
But for each $y \in \R$ we have that $\map \phi {x, y} = \map C y$ is a constant.
By smoothness of test functions, this constant has to be the same for all $y \in \R$.
Hence, $\map \phi {x, y} = 0$.
{{qed}}
Theorem: Construction of Conic Section

Consider a right circular cone $\CC$ with opening angle $2 \alpha$ whose apex is at $O$.
Consider a plane $\PP$ which intersects $\CC$, not passing through $O$, at an angle $\beta$ to the axis of $\CC$.
Then the points of intersection form one of the following:
:a circle
:an ellipse
:a hyperbola
:a parabola.

Proof:
When $\beta$ is a right angle, the points of intersection form a circle, by definition of transverse section.
{{qed|lemma}}
Otherwise, let the plane $OAA'$ through the axis of $\CC$ perpendicular to $\PP$ intersect $\PP$ in the line $AA'$.
Let $P$ be an arbitrary point on the intersection of $\PP$ with $\CC$.
Let $PM$ be constructed perpendicular to $AA'$.
$PM$ is then perpendicular to the plane $OAA'$.
The transverse section through $P$ then contains $PM$ and cuts $OAA'$ in $TT'$, which is a diameter of that transverse section.
The diagram below is the cross-section through $\CC$ corresponding to the plane $OAA'$.
The point $P$ is imagined to be perpendicular to the plane whose projection onto $OAA'$ coincides with $M$.
:360px
Because the transverse section is a circle, we have by the Intersecting Chords Theorem:
:$PM^2 = TM \cdot MT'$
Also:
:$\dfrac {TM} {AM} = \dfrac {\map \sin {\beta + \alpha} } {\cos \alpha}$
:$\dfrac {MT'} {MA'} = \dfrac {\map \sin {\beta - \alpha} } {\cos \alpha}$
{{explain|The above need to be demonstrated}}
and so:
:$PM^2 = k \cdot AM \cdot MA'$
where $k$ is the constant:
:$k = \dfrac {\map \sin {\beta + \alpha} \map \sin {\beta - \alpha} } {\cos \alpha}$
When $\beta > \alpha$, $\PP$ intersects one nappe of $\CC$.
In this case the intersection is a closed curve which is the ellipse, and $k$ is positive.
The plane passing through $O$ which is parallel to $\PP$ intersects $\CC$ at $O$ only.
{{qed|lemma}}
When $\beta < \alpha$, $\PP$ intersects both nappes of $\CC$.
In this case the intersection consists of two open curves which together form the hyperbola, and $k$ is negative.
The plane passing through $O$ which is parallel to $\PP$ intersects $\CC$ in two straight lines which themselves intersect at $O$.
{{qed|lemma}}
When $\beta = \alpha$, $\PP$ intersects one nappes of $\CC$.
In this case the intersection consists of one open curves which is the parabola, and $k$ is zero.
The plane passing through $O$ which is parallel to $\PP$ intersects $\CC$ in one straight line, which is a generatrix of $\CC$.
Hence:
:$MT' = 2 OA \sin \alpha$
which is constant.
Also:
:$TM = 2 AM \sin \alpha$
and so:
:$PM^2 = 4 OA \sin^2 \alpha \cdot AM$
That is:
:$PM^2 : AM$ is constant.
{{qed}}
Theorem: Dandelin's Theorem

Let $\CC$ be a double napped right circular cone with apex $O$.
Let $\PP$ be a plane which intersects $\CC$ such that:
:$\PP$ does not pass through $O$
:$\PP$ is not perpendicular to the axis of $\CC$.
Let $\EE$ be the conic section arising as the intersection between $\PP$ and $\CC$.
Let $\SS$ and $\SS'$ be the Dandelin spheres with respect to $\PP$.

Proof:
Let $\SS$ and $\SS'$ be the Dandelin spheres with respect to $\PP$.
Let $P$ be a point on $\EE$.
Let $F$ and $F'$ be the points at which $\SS$ and $\SS'$ are tangent to $\PP$ respectively.
Let the generatrix of $\CC$ which passes through $P$ touch $\SS$ and $\SS'$ at $E$ and $E'$ respectively.
Let $\theta$ be half the opening angle of $\CC$.
Let $\phi$ be the inclination of $\PP$ to the axis of $\CC$.
Let $\PP$ intersect $\KK$ in the straight line $NX$.
Let $PN$ be constructed perpendicular to $NX$.
Let $PK$ be constructed perpendicular to $\KK$.
Then:
:$PK = PN \cos \phi$
Also:
:$PK = PE \cos \theta = PF \cos \theta$
Hence:
:$\dfrac {PF} {PN} = \dfrac {\cos \phi} {\cos \theta}$
which is constant.
Hence $NX$ is a directrix of $\EE$ by definition.
A similar argument and construction applies with respect to $\SS'$ and $\KK$.
{{qed}}