Parabola, Ellipse & Hyperbola

LessonAbout 1.5 h2 of 2 in Analytic Geometry 1

Identification and General Equation of Conic Sections

A conic section is the locus of a point that moves such that the ratio of its distance from a fixed point (focus) to its distance from a fixed straight line (directrix) is a constant value called eccentricity (\(e\)). Conic sections are formed by the intersection of a double-napped right circular cone and a cutting plane.

General Second-Degree Equation:
\[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \]
Identification using the discriminant (\(B^2 - 4AC\)):
\[ B^2 - 4AC < 0 \implies \text{Ellipse } (e < 1.0) \]
\[ B^2 - 4AC = 0 \implies \text{Parabola } (e = 1.0) \]
\[ B^2 - 4AC > 0 \implies \text{Hyperbola } (e > 1.0) \]
When principal axes are parallel to coordinate axes (\(B = 0\)):
\[ A = C \implies \text{Circle} \]
\[ A \neq C \text{ (same signs)} \implies \text{Ellipse} \]
\[ A \text{ and } C \text{ (opposite signs)} \implies \text{Hyperbola} \]
\[ A = 0 \text{ or } C = 0 \implies \text{Parabola} \]
  • Circle results when cutting plane is parallel to base
  • Ellipse results when cutting plane is inclined to base
  • Parabola results when cutting plane is parallel to element
  • Hyperbola results when cutting plane is parallel to axis

Worked Example 1: Identifying a conic curve from its general equation

Identify the conic section described by the equation \(3x^2 + 2y^2 + 10x - 6y - 55 = 0\).

Solution: Compare the equation coefficients with the general form \(Ax^2 + Cy^2 + Dx + Ey + F = 0\). Here \(B = 0\), \(A = 3\), and \(C = 2\). Since \(A \neq C\) and both coefficients have the same positive sign, the conic is an ellipse.

The Parabola

A parabola is the locus of a point that moves so that its distance from a fixed focus is always equal to its distance from a fixed line called the directrix, giving an eccentricity \(e = 1\). The vertex \(V(h, k)\) is the turning point of the curve, located halfway between the focus and the directrix.

Standard Equation with vertex \(V(h, k)\):
\[ \text{Axis parallel to x-axis (opens right/left): } (y - k)^2 = \pm 4a(x - h) \]
\[ \text{Axis parallel to y-axis (opens up/down): } (x - h)^2 = \pm 4a(y - k) \]
Latus Rectum (\(LR\)) and Focal Length (\(a\)):
\[ LR = 4a \]
Vertex and focal length formulas for general horizontal form \(Cy^2 + Dx + Ey + F = 0\):
\[ h = \dfrac{E^2 - 4CF}{4CD}, \qquad\qquad k = -\dfrac{E}{2C}, \qquad\qquad a = -\dfrac{D}{4C} \]
Vertex and focal length formulas for general vertical form \(Ax^2 + Dx + Ey + F = 0\):
\[ h = -\dfrac{D}{2A}, \qquad\qquad k = \dfrac{D^2 - 4AF}{4AE}, \qquad\qquad a = -\dfrac{E}{4A} \]
Area bounded by parabola and its latus rectum:
\[ A_{\text{area}} = \dfrac{2}{3}(LR)a = \dfrac{8}{3}a^2 \]
  • Locus of points equidistant from focus and directrix line
  • Eccentricity of every parabola equals exactly one
  • Latus rectum length equals four times focal distance
  • Vertex lies halfway between focus and directrix line

Worked Example 2: Vertex and focal properties from standard form

Determine the vertex coordinates, focal length \(a\), and latus rectum length for the parabola \(y^2 + 8x - 6y + 25 = 0\).

Solution: Rearrange and complete the square for the \(y\) terms:

\[ y^2 - 6y = -8x - 25 \implies (y - 3)^2 = -8x - 25 + 9 \]
\[ (y - 3)^2 = -8x - 16 = -8(x + 2) \]

Comparing to \((y - k)^2 = -4a(x - h)\) yields vertex \(V(-2, 3)\), \(4a = 8 \implies a = 2\), and latus rectum length \(LR = 8\).

Worked Example 3: Directrix equation of a parabola

Find the equation of the directrix of the parabola \((y - 2)^2 = -8(x - 1)\).

Solution: The vertex is \(V(1, 2)\) and \(4a = 8 \implies a = 2\). Since the squared term is \((y - 2)^2\) with a negative sign, the parabola opens to the left. The directrix is a vertical line located \(a = 2\) units to the right of the vertex:

\[ x = h + a = 1 + 2 = 3 \]

Worked Example 4: Area bounded by parabola and latus rectum

Calculate the area enclosed by the parabola \(x^2 = 16(y - 1)\) and its latus rectum.

Solution: From \(x^2 = 16(y - 1)\), \(4a = 16 \implies a = 4\) and \(LR = 16\). Apply the parabolic area formula:

\[ A_{\text{area}} = \dfrac{2}{3}(LR)a = \dfrac{2}{3}(16)(4) = \dfrac{128}{3} \approx 42.67\text{ sq. units} \]

The Ellipse

An ellipse is the locus of a point that moves such that the sum of its distances from two fixed points (foci) is constant and equal to the length of the major axis (\(2a\)). Its eccentricity is strictly less than one (\(e < 1.0\)).

Fundamental axis relationship:
\[ a^2 = b^2 + c^2 \qquad\qquad (a > b) \]
Standard Equation with center \(C(h, k)\):
\[ \text{Major axis horizontal: } \dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1 \]
\[ \text{Major axis vertical: } \dfrac{(x - h)^2}{b^2} + \dfrac{(y - k)^2}{a^2} = 1 \]
Eccentricity (\(e\)), Directrix distance (\(d\)), and Latus Rectum (\(LR\)):
\[ e = \dfrac{c}{a} < 1.0, \qquad\qquad d = \dfrac{a}{e} = \dfrac{a^2}{c}, \qquad\qquad LR = \dfrac{2b^2}{a} \]
Area of an ellipse:
\[ A_{\text{area}} = \pi a b \]
Orbital distances from focus (Apogee and Perigee):
\[ \text{Apogee} = a + c = a(1 + e), \qquad\qquad \text{Perigee} = a - c = a(1 - e) \]
  • Sum of distances to foci equals major axis length
  • Eccentricity of an ellipse is strictly less than one
  • Major axis length always exceeds minor axis length
  • Latus rectum length equals two b squared over a

Worked Example 5: Center, semi-axes, and area of an ellipse

Find the center coordinates, semi-major axis \(a\), semi-minor axis \(b\), and enclosed area for the ellipse \(9x^2 + 25y^2 - 225 = 0\).

Solution: Divide the entire equation by \(225\) to achieve standard form:

\[ \dfrac{x^2}{25} + \dfrac{y^2}{9} = 1 \]

The center is at \(C(0, 0)\). Here \(a^2 = 25 \implies a = 5\) and \(b^2 = 9 \implies b = 3\). Calculate the area:

\[ A_{\text{area}} = \pi a b = \pi(5)(3) = 15\pi \approx 47.12\text{ sq. units} \]

Worked Example 6: Eccentricity and latus rectum calculation

An ellipse has a major axis length of \(10\text{ cm}\) and a minor axis length of \(8\text{ cm}\). Find its focal length \(c\), eccentricity \(e\), and latus rectum length \(LR\).

Solution: The semi-major axis is \(a = 5\) and semi-minor axis is \(b = 4\).

\[ c = \sqrt{a^2 - b^2} = \sqrt{5^2 - 4^2} = \sqrt{9} = 3 \]
\[ e = \dfrac{c}{a} = \dfrac{3}{5} = 0.60 \]
\[ LR = \dfrac{2b^2}{a} = \dfrac{2(4^2)}{5} = \dfrac{32}{5} = 6.4\text{ cm} \]

Worked Example 7: Astronomical apogee calculation

An Earth satellite moves in an elliptical orbit around the Earth (located at one focus). Given a semi-major axis \(a = 29,700\text{ km}\) and a focus-to-center distance \(c = 16,700\text{ km}\), determine the apogee distance.

Solution: Apogee represents the maximum distance from the focus to the orbital path:

\[ \text{Apogee} = a + c = 29,700 + 16,700 = 46,400\text{ km} \]

The Hyperbola

A hyperbola is the locus of a point that moves such that the absolute difference of its distances from two fixed foci is constant and equal to the length of the transverse axis (\(2a\)). Its eccentricity is strictly greater than one (\(e > 1.0\)).

Fundamental axis relationship:
\[ c^2 = a^2 + b^2 \]
Standard Equation with center \(C(h, k)\):
\[ \text{Transverse axis horizontal: } \dfrac{(x - h)^2}{a^2} - \dfrac{(y - k)^2}{b^2} = 1 \]
\[ \text{Transverse axis vertical: } \dfrac{(y - k)^2}{a^2} - \dfrac{(x - h)^2}{b^2} = 1 \]
Equations of Asymptotes (\(y - k = \pm m(x - h)\)):
\[ m = \pm \dfrac{b}{a} \qquad\qquad (\text{horizontal transverse axis}) \]
\[ m = \pm \dfrac{a}{b} \qquad\qquad (\text{vertical transverse axis}) \]
Eccentricity (\(e\)) and Latus Rectum (\(LR\)):
\[ e = \dfrac{c}{a} > 1.0, \qquad\qquad LR = \dfrac{2b^2}{a} \]
  • Difference of distances to foci equals transverse axis length
  • Eccentricity of a hyperbola is strictly greater than one
  • Asymptotes guide the direction of hyperbola branches
  • Conjugate axis is perpendicular to the transverse axis

Worked Example 8: Hyperbola center, semi-axes, and focal distance

Find the center, semi-transverse axis \(a\), semi-conjugate axis \(b\), and focal distance \(c\) for the hyperbola \(9x^2 - 4y^2 - 36x + 8y = 4\).

Solution: Group variables and complete the square:

\[ 9(x^2 - 4x + 4) - 4(y^2 - 2y + 1) = 4 + 36 - 4 \]
\[ 9(x - 2)^2 - 4(y - 1)^2 = 36 \]

Divide by \(36\) to achieve standard form:

\[ \dfrac{(x - 2)^2}{4} - \dfrac{(y - 1)^2}{9} = 1 \]

The center is \(C(2, 1)\). Here \(a^2 = 4 \implies a = 2\) and \(b^2 = 9 \implies b = 3\). Calculate focal distance \(c\):

\[ c = \sqrt{a^2 + b^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.61 \]

Worked Example 9: Equations of hyperbola asymptotes

Determine the equations of the asymptotes for the hyperbola \(\dfrac{x^2}{9} - \dfrac{y^2}{4} = 1\).

Solution: Center is \(C(0, 0)\), \(a = 3\), and \(b = 2\). Since the transverse axis is horizontal, slope \(m = \pm \dfrac{b}{a} = \pm \dfrac{2}{3}\).

\[ y = \pm \dfrac{2}{3}x \implies 2x - 3y = 0 \qquad\qquad \text{and} \qquad\qquad 2x + 3y = 0 \]

Polar Coordinates and Conversions

The polar coordinate system locates a point \(P(r, \theta)\) using a directed distance \(r\) (radius vector) from a fixed pole \(O\), and a polar angle \(\theta\) measured counterclockwise from a fixed horizontal polar axis.

Conversion from Polar \((r, \theta)\) to Rectangular \((x, y)\):
\[ x = r \cos\theta, \qquad\qquad y = r \sin\theta \]
Conversion from Rectangular \((x, y)\) to Polar \((r, \theta)\):
\[ r = \sqrt{x^2 + y^2}, \qquad\qquad \theta = \arctan\left(\dfrac{y}{x}\right) \]
Distance between two polar points \((r_1, \theta_1)\) and \((r_2, \theta_2)\):
\[ d = \sqrt{r_1^2 + r_2^2 - 2 r_1 r_2 \cos(\theta_2 - \theta_1)} \]
Polar equation of a circle centered at \((a, 0)\) with radius \(a\):
\[ r = 2a \cos\theta \]
  • Radius vector measures distance from pole to point
  • Polar angle is measured counterclockwise from polar axis
  • Pole corresponds to origin in rectangular coordinate system
  • Distance formula uses cosine law in polar coordinates

Worked Example 10: Rectangular to polar coordinate conversion

Convert the rectangular coordinates \((-6, -8)\) to polar coordinates \((r, \theta)\).

Solution: Compute the radius vector \(r\):

\[ r = \sqrt{(-6)^2 + (-8)^2} = \sqrt{36 + 64} = 10 \]

Find the reference angle \(\theta_{\text{ref}}\):

\[ \theta_{\text{ref}} = \arctan\left|\dfrac{-8}{-6}\right| = \arctan(1.3333) = 53.13^\circ \]

Since both \(x\) and \(y\) are negative, the point lies in Quadrant III:

\[ \theta = 180^\circ + 53.13^\circ = 233.13^\circ \]

The polar coordinates are \((10, 233.13^\circ)\).

Worked Example 11: Polar equation of a circle

Find the polar equation of a circle with radius \(4\) centered at \((4, 0)\) in rectangular coordinates.

Solution: The rectangular equation is \((x - 4)^2 + y^2 = 4^2 \implies x^2 - 8x + 16 + y^2 = 16 \implies x^2 + y^2 - 8x = 0\).

Substitute \(x^2 + y^2 = r^2\) and \(x = r \cos\theta\):

\[ r^2 - 8r \cos\theta = 0 \implies r = 8\cos\theta \qquad\qquad (\text{or } r - 8\cos\theta = 0) \]

Worked Example 12: Converting a polar equation to rectangular form

Identify the rectangular equation of the curve given by \(r \sin^2\theta = \cos\theta\).

Solution: Substitute \(\sin\theta = \dfrac{y}{r}\) and \(\cos\theta = \dfrac{x}{r}\) into the polar equation:

\[ r \left(\dfrac{y}{r}\right)^2 = \dfrac{x}{r} \implies r \cdot \dfrac{y^2}{r^2} = \dfrac{x}{r} \implies \dfrac{y^2}{r} = \dfrac{x}{r} \implies y^2 = x \]

This represents a parabola with vertex at \((0, 0)\) opening to the right.

Formula Summary: Analytic Geometry (Conic Sections and Polar Coordinates)

Formula Name Formula
General Conic Section Equation \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\)
Conic Discriminant \(B^2 - 4AC\)
Parabola Standard Equation (Horizontal Axis) \((y - k)^2 = \pm 4a(x - h)\)
Parabola Standard Equation (Vertical Axis) \((x - h)^2 = \pm 4a(y - k)\)
Parabola Latus Rectum \(LR = 4a\)
Parabola Bounded Area \(A = \dfrac{2}{3}(LR)a = \dfrac{8}{3}a^2\)
Ellipse Semi-Axes Relation \(a^2 = b^2 + c^2\)
Ellipse Standard Equation (Horizontal Major Axis) \(\dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1\)
Ellipse Eccentricity \(e = \dfrac{c}{a}\)
Ellipse Latus Rectum \(LR = \dfrac{2b^2}{a}\)
Ellipse Area \(A = \pi a b\)
Orbital Apogee and Perigee \(\text{Apogee} = a(1 + e), \quad \text{Perigee} = a(1 - e)\)
Hyperbola Semi-Axes Relation \(c^2 = a^2 + b^2\)
Hyperbola Standard Equation (Horizontal Transverse Axis) \(\dfrac{(x - h)^2}{a^2} - \dfrac{(y - k)^2}{b^2} = 1\)
Hyperbola Asymptote Slope \(m = \pm \dfrac{b}{a}\)
Hyperbola Eccentricity \(e = \dfrac{c}{a}\)
Hyperbola Latus Rectum \(LR = \dfrac{2b^2}{a}\)
Polar to Rectangular Conversion \(x = r \cos\theta, \quad y = r \sin\theta\)
Rectangular to Polar Conversion \(r = \sqrt{x^2 + y^2}, \quad \theta = \arctan\left(\dfrac{y}{x}\right)\)
Polar Distance Formula \(d = \sqrt{r_1^2 + r_2^2 - 2 r_1 r_2 \cos(\theta_2 - \theta_1)}\)
Polar Circle Equation \(r = 2a \cos\theta\)