In-Depth Study of the Parabola

1. The Focus-Directrix Locus Definition

A parabola is defined geometrically as the planar locus of a point $P(x, y)$ that moves such that its distance from a fixed point $S$ (the focus) is strictly equal to its perpendicular distance from a fixed straight line $D$ (the directrix): $$\frac{SP}{PM} = e = 1 \iff SP = PM$$ where $M$ is the foot of the perpendicular from $P$ to directrix $D$.

To derive the canonical Cartesian equation:

1. Choose the focus at $S(a, 0)$ with $a > 0$.
2. Choose the directrix as the vertical line $D: x + a = 0 \iff x = -a$.
3. The point $M$ has coordinates $(-a, y)$.
4. Equating distances: $$SP^2 = PM^2 \implies (x - a)^2 + (y - 0)^2 = (x - (-a))^2 + (y - y)^2$$ $$x^2 - 2ax + a^2 + y^2 = (x + a)^2 = x^2 + 2ax + a^2$$ Canceling $x^2 + a^2$ on both sides yields the classical canonical equation: $$\mathbf{y^2 = 4ax}$$

2. Geometric Elements of the Standard Parabola ($y^2 = 4ax$)

• Vertex ($V$): The origin $V(0, 0)$, midpoint of the perpendicular from focus to directrix.
• Axis of Symmetry: The $x$-axis ($y = 0$). The curve is symmetric about this line because $y = \pm 2\sqrt{ax}$.
• Focus ($S$): $S(a, 0)$.
• Directrix ($D$): The vertical line $x = -a$.
• Focal Distance: For any point $P(x_1, y_1)$ on the curve: $$SP = x_1 + a$$
• Latus Rectum ($LL'$): The focal chord perpendicular to the axis of symmetry. Substituting $x = a$ into $y^2 = 4ax$ yields $y^2 = 4a^2 \implies y = \pm 2a$. The extremities are $L(a, 2a)$ and $L'(a, -2a)$, and its total length is: $$\mathbf{Length(LL') = 4a}$$

3. Canonical Orientations of the Parabola

Depending on the direction of opening and orientation of the axis:

1. The Standard Rational Parametrization

The equation $y^2 = 4ax$ can be parametrized rationally without radicals by setting $y = 2at$: $$(2at)^2 = 4ax \implies 4a^2 t^2 = 4ax \implies x = at^2$$ Thus, any point on the parabola is uniquely identified by the real parameter $t \in \mathbb{R}$: $$P(t) = (a t^2, 2 a t)$$ Notice that the parameter $t$ is the reciprocal of the slope of the tangent at $P$ ($m = 1/t$).

2. Chord Joining Two Points & The Focal Chord Theorem

Let $P(t_1) = (a t_1^2, 2 a t_1)$ and $Q(t_2) = (a t_2^2, 2 a t_2)$ be two distinct points on the parabola. The slope of the secant chord $PQ$ is: $$m_{PQ} = \frac{2a t_2 - 2a t_1}{a t_2^2 - a t_1^2} = \frac{2a(t_2 - t_1)}{a(t_2 - t_1)(t_2 + t_1)} = \frac{2}{t_1 + t_2}$$ The equation of the chord $PQ$ in point-slope form is: $$y - 2a t_1 = \frac{2}{t_1 + t_2}(x - a t_1^2) \iff (t_1 + t_2)y = 2x + 2a t_1 t_2$$

The Fundamental Focal Chord Condition: If the chord $PQ$ passes through the focus $S(a, 0)$, substituting $x = a, y = 0$ yields: $$(t_1 + t_2)(0) = 2a + 2a t_1 t_2 \implies 2a(1 + t_1 t_2) = 0 \iff \mathbf{t_1 t_2 = -1 \iff t_2 = -\frac{1}{t_1}}$$ This theorem leads to two crucial geometric properties:

• Harmonic Mean Property: The semi-latus rectum $2a$ is the harmonic mean of the two focal segments $SP$ and $SQ$: $$SP = a(t_1^2 + 1), \quad SQ = a(t_2^2 + 1) = a\left(\frac{1}{t_1^2} + 1\right) = a\frac{t_1^2 + 1}{t_1^2}$$ $$\frac{1}{SP} + \frac{1}{SQ} = \frac{1}{a(t_1^2 + 1)} + \frac{t_1^2}{a(t_1^2 + 1)} = \frac{1 + t_1^2}{a(1 + t_1^2)} = \frac{1}{a} \iff \mathbf{\frac{2}{PQ_{\text{harm}}} = \frac{1}{a}}$$
• Total Length of Focal Chord: $$PQ = SP + SQ = a\left(t_1 + \frac{1}{t_1}\right)^2 \ge 4a$$ with minimum length $4a$ occurring when $t_1 = 1$ (the latus rectum).

1. The Three Canonical Forms of Tangents

For the parabola $y^2 = 4ax$:

1. Point Form: Tangent at $P(x_1, y_1)$ on the curve: $$y y_1 = 2a(x + x_1)$$
2. Parametric Form: Tangent at $P(t) = (at^2, 2at)$: $$y(2at) = 2a(x + at^2) \iff \mathbf{t y = x + a t^2}$$ The slope of this tangent is $m = 1/t$.
3. Slope Form: Replacing $t = 1/m$: $$\frac{y}{m} = x + \frac{a}{m^2} \iff \mathbf{y = m x + \frac{a}{m} \quad (m \ne 0)}$$ The point of tangency is $\left(\frac{a}{m^2}, \frac{2a}{m}\right)$.

2. Point of Intersection of Two Tangents

Let the tangents be drawn at $P(t_1)$ and $Q(t_2)$: $$t_1 y = x + a t_1^2, \qquad t_2 y = x + a t_2^2$$ Subtracting the equations: $$(t_1 - t_2)y = a(t_1^2 - t_2^2) = a(t_1 - t_2)(t_1 + t_2) \implies y = a(t_1 + t_2)$$ Substituting back into $x = t_1 y - a t_1^2$: $$x = t_1[a(t_1 + t_2)] - a t_1^2 = a t_1^2 + a t_1 t_2 - a t_1^2 = a t_1 t_2$$ Thus, the intersection of tangents at $t_1$ and $t_2$ is: $$\mathbf{T = (a t_1 t_2, \; a(t_1 + t_2))}$$ Notice: The $x$-coordinate is the geometric mean of the abscissae, and the $y$-coordinate is the arithmetic mean of the ordinates of $P$ and $Q$!

3. Orthoptic Property (The Director Circle is the Directrix)

If the tangents at $P(t_1)$ and $Q(t_2)$ are mutually perpendicular, the product of their slopes is $-1$: $$m_1 m_2 = \left(\frac{1}{t_1}\right)\left(\frac{1}{t_2}\right) = -1 \iff t_1 t_2 = -1$$ Substituting $t_1 t_2 = -1$ into the intersection coordinate: $$x_T = a(t_1 t_2) = a(-1) = -a$$ The Orthoptic Theorem: The locus of the point of intersection of two mutually perpendicular tangents to a parabola is its directrix $x = -a$! (For a parabola, the director circle degenerates into its directrix).

1. Equations of the Normal

The normal at point $P(x_1, y_1)$ on $y^2 = 4ax$ has slope $m_N = -y_1/(2a)$: $$y - y_1 = -\frac{y_1}{2a}(x - x_1)$$ In terms of the parameter $t$ ($m_N = -t$): $$y - 2at = -t(x - at^2) \iff \mathbf{y + t x = 2 a t + a t^3}$$ Setting slope $m = -t$: $$\mathbf{y = m x - 2 a m - a m^3}$$

2. The Three Co-Normal Points Theorem

If a normal passes through a given point $(\alpha, \beta)$, then: $$\beta = m \alpha - 2am - am^3 \iff a m^3 + (2a - \alpha)m + \beta = 0$$ This is a cubic polynomial in the slope $m$. By the Fundamental Theorem of Algebra, it has three roots $m_1, m_2, m_3$ (at least one of which must be real). Thus, from any point in the plane, up to three normals can be drawn to a parabola!

By Viète's formulas for $a m^3 + 0 m^2 + (2a - \alpha)m + \beta = 0$: $$m_1 + m_2 + m_3 = 0$$ $$m_1 m_2 + m_2 m_3 + m_3 m_1 = \frac{2a - \alpha}{a}$$ $$m_1 m_2 m_3 = -\frac{\beta}{a}$$ Since the ordinate of the feet of the normals is $y_i = 2am_i$ (using $m_i = -t_i$): $$y_1 + y_2 + y_3 = -2a(m_1 + m_2 + m_3) = -2a(0) = \mathbf{0}$$ Theorem: The algebraic sum of the ordinates of the feet of three co-normal points on a parabola is always identically zero! Consequently, the centroid of the triangle formed by the three co-normal points always lies strictly on the axis of the parabola.

1. The Optical Reflection Property of the Parabola

Let $P(at^2, 2at)$ be a point on the parabola $y^2 = 4ax$. Let a light ray traveling parallel to the axis of symmetry (horizontal) strike the parabolic mirror at $P$. The tangent at $P$ makes an angle $\alpha$ with the axis where $\tan \alpha = 1/t$. The vector from the focus $S(a, 0)$ to $P$ makes an angle $\theta$ with the axis: $$\tan \theta = \frac{2at}{at^2 - a} = \frac{2t}{t^2 - 1} = \tan(2\alpha)$$ Therefore, the angle of the focal ray is exactly twice the angle of the tangent ray: $\theta = 2\alpha$. This proves that the tangent line bisects the angle between the focal ray $SP$ and the horizontal incident ray! By the law of specular reflection (angle of incidence equals angle of reflection): $$\mathbf{\text{All rays parallel to the axis of symmetry reflect precisely through the focus } S!}$$ This profound property is the physical foundation of satellite dishes, solar concentrators, radio telescopes, and automotive parabolic headlamps.

2. Subtangent and Subnormal Lengths

Let the tangent and normal at $P(x_1, y_1)$ intersect the $x$-axis at $T$ and $N$ respectively, and let $M(x_1, 0)$ be the projection of $P$ on the axis:

• Subtangent ($TM$): The tangent is $y y_1 = 2a(x + x_1)$. Setting $y = 0 \implies x_T = -x_1$. Thus: $$TM = |x_1 - (-x_1)| = \mathbf{2 x_1}$$ The vertex $V(0, 0)$ is the exact midpoint of $TM$!
• Subnormal ($MN$): The normal is $y - y_1 = -\frac{y_1}{2a}(x - x_1)$. Setting $y = 0 \implies -y_1 = -\frac{y_1}{2a}(x_N - x_1) \implies x_N - x_1 = 2a$. Thus: $$MN = x_N - x_1 = \mathbf{2a = \text{Constant Everywhere!}}$$ The subnormal of a parabola is constant at all points on the curve and equal to the semi-latus rectum!