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Welcome to Conics

Samuel Dominic Chukwuemeka (SamDom For Peace)
I greet you this day,
First: read the notes.
Second: view the videos.
Third: solve the questions/solved examples for each conic.
Fourth: solve the word problems.
Fifth: check your solutions with my thoroughly-explained examples.
Sixth: check your solutions with the calculators as applicable.
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Comments, ideas, areas of improvement, questions, and constructive criticisms are welcome. You may contact me.
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Thank you for visiting.

Samuel Dominic Chukwuemeka (SamDom For Peace) B.Eng., A.A.T, M.Ed., M.S



Introduction



Circle

Circle

A Circle is defined as the locus of all points on a plane equidistant (equal distance) from a fixed point.
The fixed point is the center of the circle.
The equal distance is the radius of the circle.

The equation of a circle can be written in any of these two forms:

Standard Form
$(x - h)^2 + (y - k)^2 = r^2$
where:
$x, y$ are the variables
$(h, k)$ are the coordinates of the center of the circle
$r$ is the radius of the circle

General Form
$x^2 + y^2 + cx + dy + e = 0$
where:
$x, y$ are the variables
$c$ is the coefficient of $x$
$d$ is the coefficient of $y$
$c, d, e$ are values/constants

NOTE For:
(1.) $(x - h)^2 + (y - k)^2 = r^2$; Center = $(h, k)$, Radius = $r$
This is the Standard Form.
It is our basic reference for discussing all other forms.

(2.) $(x - h)^2 + (y + k)^2 = r^2$; Center = $(h, -k)$, Radius = $r$

(3.) $(x + h)^2 + (y - k)^2 = r^2$; Center = $(-h, k)$, Radius = $r$

(4.) $(x + h)^2 + (y + k)^2 = r^2$; Center = $(-h, -k)$, Radius = $r$

(5.) $(x - h)^2 + (y - k)^2 = e$; Center = $(h, k)$, Radius = $\sqrt{e}$

(6.) $(x - h)^2 + (y + k)^2 = e$; Center = $(h, -k)$, Radius = $\sqrt{e}$

(7.) $(x + h)^2 + (y - k)^2 = e$; Center = $(-h, k)$, Radius = $\sqrt{e}$

(8.) $(x + h)^2 + (y + k)^2 = e$; Center = $(-h, -k)$, Radius = $\sqrt{e}$

(9.) When converting from the Standard Form to the General Form, expand.
Multiply the binomials and arrange the terms in order.

(10.) When converting from the General Form to the Standard Form, the Completing the Square method is used.
Then, arrange the terms in order.

(11.) If the circle is tangent to a line, there are at least two approaches we can use to find the radius.
First Approach: Graphical Approach: The radius is the distance between the center of the circle and the tangent line
Second Approach: Algebraic Approach: The radius is the absolute value of the difference between a coordinate of the center of the circle and the respective axis of the tangent line.
It is the absolute value of a difference because it cannot have a negative value.

(12.) If the circle is tangent to the line: x = some value, then the radius is the absolute value of the difference between the x-coordinate of the center and the x-value of the line.

(13.) If the circle is tangent to the line: y = some value, then the radius is the absolute value of the difference between the y-coordinate of the center and the y-value of the line.

(14.) If the circle is tangent to the x-axis: set y = 0 because all y-values are zero on the horizontal axis (x-axis).
The radius is the absolute value of the difference between the y-coordinate of the center and zero.
This means that the radius is the absolute value of the y-coordinate of the center.

(15.) If the circle is tangent to the y-axis: set x = 0 because all x-values are zero on the vertical axis (y-axis).
The radius is the absolute value of the difference between the x-coordinate of the center and zero.
This means that the radius is the absolute value of the x-coordinate of the center.

Symbols and Meanings

  • $(h, k)$ = coordinates of the center of a circle
  • $r$ = radius of a circle
  • $d$ = diameter of a circle
  • $C$ = circumference of a circle
  • $A$ = area of a circle
  • $\pi$ = pi = $\dfrac{22}{7}$
  • $(x_1, y_1)$ = first endpoint of the diameter of a circle
  • $(x_2, y_2)$ = second endpoint of the diameter of a circle

Formulas for Circles

To solve for a specified variable for each formula, please review Solved Examples - Literal Equations

(1.) Radius, Diameter, Circumference, Area

$ d = 2r \\[3ex] r = \dfrac{d}{2} \\[5ex] C = \pi d \\[3ex] C = 2\pi r \\[3ex] A = \pi r^2 \\[3ex] A = \dfrac{\pi d^2}{4} \\[5ex] $ (2.) Standard Form of the Equation of a Circle

$(x - h)^2 + (y - k)^2 = r^2$
where:
$x, y$ are the variables
$(h, k)$ are the coordinates of the center of the circle
$r$ is the radius of the circle

(3.) General Form of the Equation of a Circle
$x^2 + y^2 + 2gx + 2fy + c = 0$
where:
$x, y$ are the variables
$c$ is the coefficient of $x$
$d$ is the coefficient of $y$
$c, d, e$ are values/constants

(4.) Converting Between Forms

Given: Standard Form of the Equation of a Circle
To Find: General Form of the Equation of the Circle

$ \underline{General\;\;Form} \\[3ex] x^2 + y^2 + 2gx + 2fy + c = 0 \\[5ex] \underline{Standard\;\;Form} \\[3ex] (x - h)^2 + (y - k)^2 = r^2 \\[3ex] (x - h)^2 + (y - k)^2 - r^2 = 0 \\[3ex] [(x - h)(x - h)] + [(y - k)(y - k)] - r^2 = 0 \\[3ex] [x^2 - hx - hx + h^2] + [y^2 - ky - ky + k^2] - r^2 = 0 \\[3ex] (x^2 - 2hx + h^2) + (y^2 - 2ky + k^2) - r^2 = 0 \\[3ex] x^2 - 2hx + h^2 + y^2 - 2ky + k^2 - r^2 = 0 \\[3ex] x^2 + y^2 - 2hx - 2ky + h^2 + k^2 - r^2 = 0 \\[5ex] General\;\;Form = Standard\;\;Form \\[3ex] \implies \\[3ex] RHS = RHS:\;\;0 = 0 ...okay \\[3ex] LHS = LHS: \\[3ex] x^2 + y^2 + 2gx + 2fy + c = x^2 + y^2 - 2hx - 2ky + h^2 + k^2 - r^2 \\[3ex] 2gx + 2fy + c = -2hx - 2ky + h^2 + k^2 - r^2 \\[3ex] \underline{Equate\;\;the\;\;terms\;\;in\;\;x} \\[3ex] 2gx = -2hx \\[3ex] g = -h \\[3ex] \underline{Equate\;\;the\;\;terms\;\;in\;\;y} \\[3ex] 2fy = -2ky \\[3ex] f = -k \\[3ex] \underline{Equate\;\;the\;\;constants} \\[3ex] c = h^2 + k^2 - r^2 \\[5ex] $ Given: General Form of the Equation of a Circle
To Find: Standard Form of the Equation of the Circle

$ \underline{Standard\;\;Form} \\[3ex] (x - h)^2 + (y - k)^2 = r^2 \\[3ex] (x - h)^2 + (y - k)^2 - r^2 = 0 \\[3ex] [(x - h)(x - h)] + [(y - k)(y - k)] - r^2 = 0 \\[3ex] [x^2 - hx - hx + h^2] + [y^2 - ky - ky + k^2] - r^2 = 0 \\[3ex] (x^2 - 2hx + h^2) + (y^2 - 2ky + k^2) - r^2 = 0 \\[3ex] x^2 - 2hx + h^2 + y^2 - 2ky + k^2 - r^2 = 0 \\[3ex] x^2 + y^2 - 2hx - 2ky + h^2 + k^2 - r^2 = 0 \\[5ex] \underline{General\;\;Form} \\[3ex] x^2 + y^2 + 2gx + 2fy + c = 0 \\[5ex] Standard\;\;Form = General\;\;Form \\[3ex] \implies \\[3ex] RHS = RHS:\;\;0 = 0 ...okay \\[3ex] LHS = LHS: \\[3ex] x^2 + y^2 - 2hx - 2ky + h^2 + k^2 - r^2 = x^2 + y^2 + 2gx + 2fy + c \\[3ex] - 2hx - 2ky + h^2 + k^2 - r^2 = 2gx + 2fy + c \\[3ex] \underline{Equate\;\;the\;\;terms\;\;in\;\;x} \\[3ex] -2hx = 2gx \\[3ex] -h = g \\[3ex] h = -g \\[3ex] \underline{Equate\;\;the\;\;terms\;\;in\;\;y} \\[3ex] -2ky = 2fy \\[3ex] -k = f \\[3ex] k = -f \\[3ex] \underline{Equate\;\;the\;\;constants} \\[3ex] h^2 + k^2 - r^2 = c \\[3ex] h^2 + k^2 - c = r^2 \\[3ex] r^2 = h^2 + k^2 - c \\[3ex] r^2 = (-g)^2 + (-f)^2 - c \\[3ex] r^2 = g^2 + f^2 - c \\[3ex] r = \sqrt{g^2 + f^2 - c} \\[5ex] $ (5.) Given: The Center Coordinates of a Circle and an Endpoint on the Circumference of the Circle
The coordinates of the center of the circle = $(h, k)$
The endpoint on the circumference of the circle = $(x_1, y_1)$
The radius of the circle can be found by the Distance Formula
The radius of the circle = $r$
r = $\sqrt{(x_1 - h)^2 + (y_1 - k)^2}$
The diameter of the circle = $d$
The diameter of the circle is twice the radius.
$d = 2 * r$
The second endpoint of the diameter of the circle can also be found
The second endpoint of the diameter of the circle = $(x_2, y_2)$

$ x_2 = x_1 + r \\[3ex] y_2 = y_1 + r \\[3ex] (x_2, y_2) = (x_1 + r, y_1 + r) \\[3ex] $ (6.) Given: The Endpoints of the Diameter of the Circle
$(x_1, y_1)$ = first endpoint of the diameter of a circle
$(x_2, y_2)$ = second endpoint of the diameter of a circle
The center of the circle is found using the Midpoint Formula
$(h, k)$ are the coordinates of the center of the circle

$ h = \dfrac{x_1 + x_2}{2} \\[5ex] k = \dfrac{y_1 + y_2}{2} \\[5ex] $ The radius of the circle can then be found by the Distance Formula using either endpoints
Please refer to Number (3.)

(7.) Given: Any Two Points
To Find: A Point on the $y-axis$ Equidistant From the Two Points
Let the first point = $(x_1, y_1)$
and the second point = $(x_2, y_2)$
A point on the $y-axis$ equidistant from $(x_1, y_1)$ and $(x_2, y_2)$ is $(0, y)$
This implies that the distance from $(x_1, y_1)$ to $(0, y)$ should be the same distance from $(0, y)$ to $(x_2, y_2)$

Using the Distance Formula to find the distance from $(x_1, y_1)$ to $(0, y)$ gives:

$ distance1 = \sqrt{(0 - x_1)^2 + (y - y_1)^2} \\[3ex] distance1 = \sqrt{(-x_1)^2 + (y - y_1)^2} \\[3ex] distance1 = \sqrt{x_1^2 + (y - y_1)^2} \\[3ex] $ Using the Distance Formula to find the distance from $(0, y)$ to $(x_2, y_2)$ gives:

$ distance2 = \sqrt{(x_2 - 0)^2 + (y_2 - y)^2} \\[3ex] distance2 = \sqrt{(x_2)^2 + (y_2 - y)^2} \\[3ex] distance2 = \sqrt{x_2^2 + (y_2 - y)^2} \\[3ex] $ The two distances should be the same ...the word, equidistant means equal distance
So, distance 1 = distance 2

$ \sqrt{x_1^2 + (y - y_1)^2} = \sqrt{x_2^2 + (y_2 - y)^2} \\[3ex] Square\:\:both\:\:sides \\[3ex] \left(\sqrt{x_1^2 + (y - y_1)^2}\right)^2 = \left(\sqrt{x_2^2 + (y_2 - y)^2}\right)^2 \\[3ex] x_1^2 + (y - y_1)^2 = x_2^2 + (y_2 - y)^2 \\[3ex] (y - y_1)^2 - (y_2 - y)^2 = x_2^2 - x_1^2 \\[3ex] Expand \\[3ex] (y - y_1)(y - y_1) - [(y_2 - y)(y_2 - y)] = x_2^2 - x_1^2 \\[3ex] y^2 - yy_1 - yy_1 + y_1^2 - (y_2^2 - yy_2 - yy_2 + y^2) = x_2^2 - x_1^2 \\[3ex] y^2 - 2yy_1 + y_1^2 - (y_2^2 - 2yy_2 + y^2) = x_2^2 - x_1^2 \\[3ex] y^2 - 2yy_1 + y_1^2 - y_2^2 + 2yy_2 - y^2 = x_2^2 - x_1^2 \\[3ex] y^2 \:\:cancels\:\:out \\[3ex] Collect\:\:like\:\:terms\:\:in\:\:y \\[3ex] 2yy_2 - 2yy_1 = x_2^2 - x_1^2 - y_1^2 + y_2^2 \\[3ex] y(2y_2 - 2y_1) = x_2^2 + y_2^2 - x_1^2 - y_1^2 \\[3ex] y = \dfrac{x_2^2 + y_2^2 - x_1^2 - y_1^2}{2y_2 - 2y_1} \\[5ex] $ (8.) Given: Any Two Points
To Find: A Point on the $x-axis$ Equidistant From the Two Points
Let the first point = $(x_1, y_1)$
and the second point = $(x_2, y_2)$
A point on the $x-axis$ equidistant from $(x_1, y_1)$ and $(x_2, y_2)$ is $(x, 0)$
This implies that the distance from $(x_1, y_1)$ to $(x, 0)$ should be the same distance from $(x, 0)$ to $(x_2, y_2)$

Using the Distance Formula to find the distance from $(x_1, y_1)$ to $(x, 0)$ gives:

$ distance1 = \sqrt{(x - x_1)^2 + (0 - y_1)^2} \\[3ex] distance1 = \sqrt{(x - x_1)^2 + (-y_1)^2} \\[3ex] distance1 = \sqrt{(x - x_1)^2 + y_1^2} \\[3ex] $ Using the Distance Formula to find the distance from $(x, 0)$ to $(x_2, y_2)$ gives:

$ distance2 = \sqrt{(x_2 - x)^2 + (y_2 - 0)^2} \\[3ex] distance2 = \sqrt{(x_2 - x)^2 + (y_2)^2} \\[3ex] distance2 = \sqrt{(x_2 - x)^2 + y_2^2} \\[3ex] $ The two distances should be the same ...the word, equidistant means equal distance
So, distance 1 = distance 2

$ \sqrt{(x - x_1)^2 + y_1^2} = \sqrt{(x_2 - x)^2 + y_2^2} \\[3ex] Square\:\:both\:\:sides \\[3ex] \left(\sqrt{(x - x_1)^2 + y_1^2}\right)^2 = \left(\sqrt{(x_2 - x)^2 + y_2^2}\right)^2 \\[3ex] (x - x_1)^2 + y_1^2 = (x_2 - x)^2 + y_2^2 \\[3ex] (x - x_1)^2 - (x_2 - x)^2 = y_2^2 - y_1^2 \\[3ex] Expand \\[3ex] (x - x_1)(x - x_1) - [(x_2 - x)(x_2 - x)] = y_2^2 - y_1^2 \\[3ex] x^2 - xx_1 - xx_1 + x_1^2 - (x_2^2 - xx_2 - xx_2 + x^2) = y_2^2 - y_1^2 \\[3ex] x^2 - 2xx_1 + x_1^2 - (x_2^2 - 2xx_2 + x^2) = y_2^2 - y_1^2 \\[3ex] x^2 - 2xx_1 + x_1^2 - x_2^2 + 2xx_2 - x^2 = y_2^2 - y_1^2 \\[3ex] x^2 \:\:cancels\:\:out \\[3ex] Collect\:\:like\:\:terms\:\:in\:\:x \\[3ex] 2xx_2 - 2xx_1 = y_2^2 - y_1^2 - x_1^2 + x_2^2 \\[3ex] x(2x_2 - 2x_1) = x_2^2 + y_2^2 - x_1^2 - y_1^2 \\[3ex] x = \dfrac{x_2^2 + y_2^2 - x_1^2 - y_1^2}{2x_2 - 2x_1} \\[5ex] $ (9.) Given: Center and Tangent to the Line Value
To Find: Radius

$ Center = (h, k) \\[3ex] \underline{Tangent\;\;to\;\;the\;\;line:\;\;x = some\;\;value} \\[3ex] r = |h - xValue| \\[3ex] On\;\;the\;\;y-axis;\;\;all\;\;x-values\;\;are\;\;0 \\[3ex] Tangent\;\;to\;\;the\;\;y-axis:\;\; set\;\; x = 0 \\[3ex] \underline{Tangent\;\;to\;\;the\;\;line:\;\;y = some\;\;value} \\[3ex] r = |k - yValue| \\[3ex] On\;\;the\;\;x-axis;\;\;all\;\;y-values\;\;are\;\;0 \\[3ex] Tangent\;\;to\;\;the\;\;x-axis:\;\; set\;\; y = 0 \\[5ex] $ (10.) Given: Any Three Points
To Find: Center Coordinates, Radius, Equation

NOTE: To do this kind of question, the knowledge of 3 * 3 Linear Systems is important.

$ 1st\;\;Point = (x_1, y_1) \\[3ex] 2nd\;\;Point = (x_2, y_2) \\[3ex] 3rd\;\;Point = (x_3, y_3) \\[3ex] \underline{General\;\;Form\;\;of\;\;the\;\;Equation\;\;of\;\;a\;\;Circle} \\[3ex] x^2 + y^2 + 2gx + 2fy + c = 0 \\[3ex] $



Ellipse

Ellipse

An Ellipse is the locus of of all the points on a plane such that the sum of the distances from two fixed points on the plane is constant.
These are some real-world examples of ellipses: the shape of an egg and the shape of the earth's orbit among others.
The two fixed points are known as the foci
The line containing the foci is known as the major axis
The eccentricity of an ellipse is between 0 and 1 (greater than zero but less than one)
The Latus Rectum of an ellipse is the line through the focus, that is perpendicular to the transverse axis.
It can also be defined as the focal chord that is parallel to the directrix of the ellipse.

Symbols and Meanings

  • (h, k) = coordinates of the center of an ellipse
  • Ellipse: a = horizontal distance from the center to the boundary
  • Ellipse: b = vertical distance from the center to the boundary
  • Horizontal Ellipse: a = half the length of the major axis OR the length of the semi-major axis
  • Horizontal Ellipse: b = half the length of the minor axis OR the length of the semi-minor axis
  • Vertical Ellipse: a = half the length of the minor axis OR the length of the semi-minor axis
  • Vertical Ellipse: b = half the length of the major axis OR the length of the semi-major axis
  • c = linear eccentricity = distance from the center of the ellipse to the foci
  • $f_1,f_2$ = foci points
  • $f$ = distance between the foci
  • e = eccentricity
  • A = area
  • P = perimeter
  • L = length of the latus rectum
  • $L_1,\;L_2,\;L_3,\;L_4$ = endpoints of the latus rectum
  • D = distance between the two directrixes

Formulas for Ellipses

To solve for a specified variable for each formula, please review Solved Examples - Literal Equations

(1.) Standard Form of the Equation of an Ellipse

$ \dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1 \\[5ex] $ If $a \gt b$, the ellipse is a horizontal ellipse.
If $a \lt b$, the ellipse is a vertical ellipse.

(2.) Area

$ A = \pi ab \\[5ex] $ (3.) Eccentricity

$ e = \dfrac{\sqrt{semiMajor\;\;axis\;\;length^2 - semiMinor\;\;axis\;\;length^2}}{semiMajor\;\;axis\;\;length} \\[3ex] $ In other words:
Eccentricity: Horizontal Ellipse:
Major axis is the horizontal axis
Semi-major axis length = a

$ e = \dfrac{\sqrt{a^2 - b^2}}{a} \\[3ex] $ Eccentricity: Vertical Ellipse:
Major axis is the vertical axis
Semi-major axis length = b

$ e = \dfrac{\sqrt{b^2 - a^2}}{b} \\[5ex] $ (4.) Foci
$foci = (f_1,f_2)$
There are at least two approaches that we can use to determine the foci of an ellipse.

Foci: Horizontal Ellipse:
Major axis is the horizontal axis

$ \underline{First\;\;Approach:\;\;Based\;\;on\;\;Linear\;\;Eccentricity\;\;and\;\;Center} \\[3ex] c^2 = a^2 - b^2 \\[3ex] c = \sqrt{a^2 - b^2} \\[3ex] Center = (h, k) \\[3ex] f_1 = (h - c, k) \\[3ex] f_2 = (h + c, k) \\[5ex] \underline{Second\;\;Approach:\;\;Based\;\;on\;\;Eccentricity\;\;and\;\;Center} \\[3ex] e = \dfrac{\sqrt{a^2 - b^2}}{a} \\[3ex] f_1 = (-ae, k) \\[3ex] f_2 = (ae, k) \\[3ex] $ Foci: Vertical Ellipse:
Major axis is the vertical axis

$ \underline{First\;\;Approach:\;\;Based\;\;on\;\;Linear\;\;Eccentricity\;\;and\;\;Center} \\[3ex] c^2 = b^2 - a^2 \\[3ex] c = \sqrt{b^2 - a^2} \\[3ex] Center = (h, k) \\[3ex] f_1 = (h, k - c) \\[3ex] f_2 = (h, k + c) \\[3ex] \underline{Second\;\;Approach:\;\;Based\;\;on\;\;Eccentricity\;\;and\;\;Center} \\[3ex] e = \dfrac{\sqrt{b^2 - a^2}}{b} \\[3ex] f_1 = (h, -be) \\[3ex] f_2 = (h, be) \\[5ex] $ (5.) Latus Rectum
There are at least two approaches that we can use to determine the endpoints of the latus rectum of an ellipse.

Latus Rectum: Horizontal Ellipse:
Major axis is the horizontal axis

$ \underline{Length} \\[3ex] L = \dfrac{2b^2}{a} \\[7ex] \underline{First\;\;Approach:\;\;Based\;\;on\;\;Linear\;\;Eccentricity\;\;and\;\;Axis} \\[3ex] c^2 = a^2 - b^2 \\[3ex] c = \sqrt{a^2 - b^2} \\[3ex] Center = (h, k) \\[3ex] L_1 = \left(h - c, -\dfrac{b^2}{a}\right) \\[5ex] L_2 = \left(h - c, \dfrac{b^2}{a}\right) \\[5ex] and \\[3ex] L_3 = \left(h + c, -\dfrac{b^2}{a}\right) \\[5ex] L_4 = \left(h + c, \dfrac{b^2}{a}\right) \\[5ex] \underline{Second\;\;Approach:\;\;Based\;\;on\;\;Eccentricity\;\;and\;\;Axis} \\[3ex] e = \dfrac{\sqrt{a^2 - b^2}}{a} \\[3ex] L_1 = \left(-ae, -\dfrac{b^2}{a}\right) \\[5ex] L_2 = \left(-ae, \dfrac{b^2}{a}\right) \\[5ex] and \\[3ex] L_3 = \left(ae, -\dfrac{b^2}{a}\right) \\[5ex] L_4 = \left(ae, \dfrac{b^2}{a}\right) \\[5ex] $ Latus Rectum: Vertical Ellipse:
Major axis is the vertical axis

$ \underline{Length} \\[3ex] L = \dfrac{2a^2}{b} \\[7ex] \underline{First\;\;Approach:\;\;Linear\;\;Eccentricity\;\;and\;\;Center} \\[3ex] c^2 = b^2 - a^2 \\[3ex] c = \sqrt{b^2 - a^2} \\[3ex] Center = (h, k) \\[3ex] L_1 = \left(-\dfrac{a^2}{b}, k - c \right) \\[5ex] L_2 = \left(\dfrac{a^2}{b}, k - c \right) \\[5ex] and \\[3ex] L_3 = \left(-\dfrac{a^2}{b}, k + c \right) \\[5ex] L_4 = \left(\dfrac{a^2}{b}, k + c \right) \\[5ex] \underline{Second\;\;Approach:\;\;Eccentricity\;\;and\;\;Center} \\[3ex] e = \dfrac{\sqrt{b^2 - a^2}}{b} \\[3ex] L_1 = \left(-\dfrac{a^2}{b}, -be \right) \\[5ex] L_2 = \left(\dfrac{a^2}{b}, -be \right) \\[5ex] and \\[3ex] L_3 = \left(-\dfrac{a^2}{b}, be \right) \\[5ex] L_4 = \left(\dfrac{a^2}{b}, be \right) \\[5ex] $ (6.) Directrix: Horizontal Ellipse:
Major axis is the horizontal axis

$ The\;\;two\;\;directrixes\;\;are: \\[3ex] x = -\dfrac{a}{e} \\[5ex] x = \dfrac{a}{e} \\[5ex] $ Horizontal Distance Between the Two Directrixes are:

$ D = \dfrac{2a}{e} \\[5ex] $ Directrix: Vertical Ellipse:
Major axis is the vertical axis

$ The\;\;two\;\;directrixes\;\;are: \\[3ex] y = -\dfrac{b}{e} \\[5ex] y = \dfrac{b}{e} \\[5ex] $ Vertical Distance Between the Two Directrixes are:

$ D = \dfrac{2b}{e} \\[5ex] $ (7.) Factored Form of the Equation of an Ellipse

$ b^2(x - h)^2 + a^2(y - k)^2 = a^2b^2 \\[5ex] $ If $a \gt b$, the ellipse is a horizontal ellipse.
If $a \lt b$, the ellipse is a vertical ellipse.

(8.) General Form of the Equation of an Ellipse

$ b^2\boldsymbol{x^2} + a^2\boldsymbols{y^2} - 2b^2h\boldsymbol{x} - 2a^2k\boldsymbol{y} + b^2h^2 + a^2k^2 - a^2b^2 = 0 \\[5ex] OR \\[3ex] b^2\boldsymbol{x^2} + a^2\boldsymbols{y^2} - 2b^2h\boldsymbol{x} - 2a^2k\boldsymbol{y} = a^2b^2 - b^2h^2 - a^2k^2 \\[5ex] $ If $a \gt b$, the ellipse is a horizontal ellipse.
If $a \lt b$, the ellipse is a vertical ellipse.



Parabola

Parabola

Symbols and Meanings

Formulas for Parabolas

To solve for a specified variable for each formula, please review Solved Examples - Literal Equations



Hyperbola

Hyperbola

Symbols and Meanings

Formulas for Hyperbolas

To solve for a specified variable for each formula, please review Solved Examples - Literal Equations



References

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