Conic sections (or conics) are curves obtained by intersecting a double-napped right circular cone with a plane. These curves have extensive applications in planetary motion, telescope design, reflectors, and acoustics.
Let $l$ be a fixed vertical line (the axis) and $m$ be another line (the generator) intersecting $l$ at a fixed point $V$ (the vertex) at an angle $\alpha$ (the semi-vertical angle). Rotating $m$ around $l$ while keeping $\alpha$ constant generates a double-napped right circular hollow cone consisting of an upper nappe and a lower nappe meeting at $V$.
Let $\beta$ be the angle made by the intersecting plane with the vertical axis $l$ of the cone. Depending on the value of $\beta$, different curves are formed:
| Section Type | Angle Condition | Geometric Description | Standard Equation |
|---|---|---|---|
| Circle | $\beta = 90^\circ$ | The plane is perpendicular to the axis and cuts across a single nappe. | $x^2 + y^2 = r^2$ |
| Ellipse | $\alpha < \beta < 90^\circ$ | The plane cuts obliquely across one entire nappe. | $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ |
| Parabola | $\beta = \alpha$ | The plane is parallel to the generator line and cuts only one nappe. | $y^2 = 4ax$ |
| Hyperbola | $0 \le \beta < \alpha$ | The plane is parallel to the axis, cutting through both the upper and lower nappes. | $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ |
When the intersecting plane passes directly through the vertex $V$ of the cone, the resulting sections are called degenerate conics:
A circle is the set of all points in a plane that are equidistant from a fixed point (the centre) in the plane. The constant distance is called the radius.
The equation of a circle with centre $(h, k)$ and radius $r$ is derived using the distance formula:
$$(x - h)^2 + (y - k)^2 = r^2$$If the centre is at the origin $(0, 0)$, the equation simplifies to:
$$x^2 + y^2 = r^2$$Mathematical Problem Statement: Find the centre and radius of the circle given by the equation $x^2 + y^2 + 8x + 10y - 8 = 0$.
Step-by-Step Algorithmic Solution Trace:
- Group the $x$ and $y$ terms: $(x^2 + 8x) + (y^2 + 10y) = 8$.
- Complete the square for $x$: add $(\frac{8}{2})^2 = 16$ to both sides.
- Complete the square for $y$: add $(\frac{10}{2})^2 = 25$ to both sides.
- Rewrite the equation: $(x^2 + 8x + 16) + (y^2 + 10y + 25) = 8 + 16 + 25$.
- Simplify: $(x + 4)^2 + (y + 5)^2 = 49$.
- Express in standard form: $(x - (-4))^2 + (y - (-5))^2 = 7^2$.
Final Evaluation Output: Centre is $(-4, -5)$ and Radius is $7$.
A parabola is the set of all points in a plane that are equidistant from a fixed line (the directrix) and a fixed point (the focus) not on the line.
The simplest form of a parabola has its vertex at the origin $(0,0)$ and its axis along one of the coordinate axes. There are four standard orientations:
| Equation | Focus | Directrix | Axis of Symmetry | Direction of Opening |
|---|---|---|---|---|
| $y^2 = 4ax$ | $(a, 0)$ | $x = -a$ | $y = 0$ ($x$-axis) | Opens Right |
| $y^2 = -4ax$ | $(-a, 0)$ | $x = a$ | $y = 0$ ($x$-axis) | Opens Left |
| $x^2 = 4ay$ | $(0, a)$ | $y = -a$ | $x = 0$ ($y$-axis) | Opens Upward |
| $x^2 = -4ay$ | $(0, -a)$ | $y = a$ | $x = 0$ ($y$-axis) | Opens Downward |
The latus rectum of a parabola is a line segment perpendicular to the axis of the parabola, passing through the focus, with endpoints lying on the parabola. For any standard parabola, the length of the latus rectum is:
$$\text{Length} = 4a$$Mathematical Problem Statement: Find the coordinates of the focus, the equation of the directrix, and the length of the latus rectum for the parabola $y^2 = 8x$.
Step-by-Step Algorithmic Solution Trace:
- Compare the given equation $y^2 = 8x$ with the standard form $y^2 = 4ax$.
- Solve for $a$: $4a = 8 \implies a = 2$.
- Determine the focus: Since the axis is along the $x$-axis and $a > 0$, the focus is $(a, 0) = (2, 0)$.
- Determine the directrix: The equation of the directrix is $x = -a \implies x = -2$.
- Calculate the length of the latus rectum: $\text{Length} = 4a = 8$.
Final Evaluation Output: Focus: $(2, 0)$, Directrix: $x = -2$, Latus Rectum Length: $8$.
An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points (the foci) in the plane is a constant ($2a$).
Let $2a$ be the length of the major axis, $2b$ be the length of the minor axis, and $2c$ be the distance between the foci. The fundamental relationship between these parameters is:
$$a^2 = b^2 + c^2 \implies c = \sqrt{a^2 - b^2}$$The eccentricity ($e$) of an ellipse is the ratio of the distance of a focus from the centre to the length of the semi-major axis:
$$e = \frac{c}{a} \quad (0 < e < 1)$$| Standard Equation | Major Axis Along | Foci Coordinates | Vertices Coordinates | Latus Rectum Length |
|---|---|---|---|---|
| $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad (a > b)$ | $x$-axis | $(\pm c, 0)$ | $(\pm a, 0)$ | $\frac{2b^2}{a}$ |
| $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \quad (a > b)$ | $y$-axis | $(0, \pm c)$ | $(0, \pm a)$ | $\frac{2b^2}{a}$ |
Mathematical Problem Statement: Find the coordinates of the foci, the vertices, the eccentricity, and the length of the latus rectum of the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$.
Step-by-Step Algorithmic Solution Trace:
- Identify $a^2$ and $b^2$ by comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$: $a^2 = 25 \implies a = 5$, and $b^2 = 9 \implies b = 3$.
- Since $a^2 > b^2$ is under $x^2$, the major axis is along the $x$-axis.
- Calculate $c$: $c = \sqrt{a^2 - b^2} = \sqrt{25 - 9} = \sqrt{16} = 4$.
- Determine foci: $(\pm c, 0) = (\pm 4, 0)$.
- Determine vertices: $(\pm a, 0) = (\pm 5, 0)$.
- Calculate eccentricity: $e = \frac{c}{a} = \frac{4}{5} = 0.8$.
- Calculate latus rectum length: $\frac{2b^2}{a} = \frac{2(9)}{5} = \frac{18}{5} = 3.6$.
Final Evaluation Output: Foci: $(\pm 4, 0)$, Vertices: $(\pm 5, 0)$, Eccentricity: $0.8$, Latus Rectum Length: $3.6$.
A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points (the foci) in the plane is a constant ($2a$).
Let $2a$ be the length of the transverse axis, $2b$ be the length of the conjugate axis, and $2c$ be the distance between the foci. The fundamental relationship is:
$$c^2 = a^2 + b^2 \implies b = \sqrt{c^2 - a^2}$$The eccentricity ($e$) of a hyperbola is:
$$e = \frac{c}{a} \quad (e > 1)$$| Standard Equation | Transverse Axis Along | Foci Coordinates | Vertices Coordinates | Latus Rectum Length |
|---|---|---|---|---|
| $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ | $x$-axis | $(\pm c, 0)$ | $(\pm a, 0)$ | $\frac{2b^2}{a}$ |
| $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ | $y$-axis | $(0, \pm c)$ | $(0, \pm a)$ | $\frac{2b^2}{a}$ |
An equilateral hyperbola is a special case where $a = b$, simplifying the equation to $x^2 - y^2 = a^2$.
Mathematical Problem Statement: Find the coordinates of the foci, the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $\frac{x^2}{9} - \frac{y^2}{16} = 1$.
Step-by-Step Algorithmic Solution Trace:
- Compare with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: $a^2 = 9 \implies a = 3$, and $b^2 = 16 \implies b = 4$.
- Calculate $c$: $c = \sqrt{a^2 + b^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
- Determine foci: $(\pm c, 0) = (\pm 5, 0)$.
- Determine vertices: $(\pm a, 0) = (\pm 3, 0)$.
- Calculate eccentricity: $e = \frac{c}{a} = \frac{5}{3}$.
- Calculate latus rectum length: $\frac{2b^2}{a} = \frac{2(16)}{3} = \frac{32}{3}$.
Final Evaluation Output: Foci: $(\pm 5, 0)$, Vertices: $(\pm 3, 0)$, Eccentricity: $\frac{5}{3}$, Latus Rectum Length: $\frac{32}{3}$.
Historical Chronicle: Apollonius of Perga (262 B.C. - 190 B.C.)
Historical Chronicle: René Descartes & Pierre de Fermat