PRE CALCULUS PARABOLA, Schemes and Mind Maps of Mathematics

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Parabola

Week 2

Learning Outcomes

The learner …

 Illustrate a parabola and its degenerate form as intersections of a plane and a two- napped cone;  define a parabola;  determine the standard form of equation of a parabola;  sketch the graph of a parabola in a rectangular coordinate system;  determine the equation of a parabola given some properties; and  solve situational problems involving a parabola.

Definitions

1. Parabola - is the set of all points in a plane equidistant from both a fixed point and a fixed line. 2. Vertex – point found in the intersection of a conic and its principal axis. 3. Latus Rectum – is the chord through the focus, and parallel to the directrix. 4. Axis of symmetry - the vertical line that goes through the vertex of a parabola. 5. Focus – is a fixed point on the interior of the parabola. 6. Directrix - a given line not through the focus.

Introduction: The Parabola

The Olympic torch concludes its journey around the world when it is used to light the Olympic cauldron during the opening ceremony. (credit: Ken Hackman, U.S. Air Force)

Did you know that the Olympic torch is lit several months before the start of the games? The ceremonial method for lighting the flame is the same as in ancient times. The ceremony takes place at the Temple of Hera in Olympia, Greece, and is rooted in Greek mythology, paying tribute to Prometheus, who stole fire from Zeus to give to all humans. One of eleven acting

 if the parabola has horizontal principal axis then, the parabola opens to the right or to the left  this line divides the parabola into two parts which are mirror images of each other.

5. Directed Distance  distance with direction  denoted by 𝑝  directed distance from the vertex to the focus  distance can be positive or negative because the sign indicates direction.  if the directed distance is positive, the parabola opens either upward or to the right  if the directed distance is negative, the parabola opens either downward or to the left
6. Latus Rectum  the length of the focal width or the focal length

Standard Equation of the Parabola:

CASE 1: Standard Equation of a Parabola that has a Vertical Principal Axis

Standard equation Vertex (^) Parabola Opens

(𝒙 − 𝒉)𝟐^ = 𝟒𝒑(𝒚 − 𝒌) Vertex at^ (ℎ,^ 𝑘)^ Upward

𝒙𝟐^ = 𝟒𝒑𝒚 Vertex at the origin (0,0)^ Downward

CASE 2: Standard Equation of a Parabola that has a Horizontal Principal Axis

Standard equation Vertex Parabola opens

(𝒚 − 𝒌)𝟐^ = 𝟒𝒑(𝒙 − 𝒉) Vertex at^ (ℎ,^ 𝑘)^ To the right

𝒚𝟐^ = 𝟒𝒑𝒙 Vertex at the origin (0,0)^ To the left

Note: The opening of the parabola depends on both the principal axis and the directed distance. Here’s the guide on how to identify where the parabola opens.

Principal Axis Vertical Horizontal Directed Distance 𝒑

𝑝 > 0 or positive upward right 𝑝 < 0 or negative downward left

Step 2: Identify the principal axis.

Since the given standard equation belongs to CASE 1 therefore, it has a vertical principal axis. From here, you can already imagine the graph of the parabola it’s either opens upward or downward.

Step 3: Solve for the directed distance (𝑝).

Since 𝒑 is positive, the parabola opens upward.

Step 4: Locate the focus by counting 𝑝 number of units from the vertex.

To locate the focus count 3 units going up because 𝑝 = 3 and the parabola has vertical principal axis.

The focus is 𝑭(𝟐, 𝟐)

Step 5: Locate the directrix. (note: the distance of the focus and directrix from the vertex is equal but the direction is opposite) To locate the directrix, count 3 units going down.

The directrix is 𝒚 = −𝟒.

Step 6 : Solve for the length of the Latus Rectum: 𝐿𝑅 = |4𝑝|

𝐿𝑅 = | 4 𝑝|

𝐿𝑅 = | 4 ( 3 )|

𝐿𝑅 = | 12 |

Step 7 : Find the endpoints of the Latus Rectum.

Clearly, the length of the latus rectum is 12 units. From the focus count 𝐿𝑅 2 = 6 units to the right to locate the 1st^ endpoint 𝐴(8,2), count another 6 units to the left from the focus to locate the 2nd^ endpoint 𝐵(−4,2). The distance from 𝐴 𝑡𝑜 𝐵 is 12 units.

Step 3: Solve for the directed distance (𝑝).

𝟓 𝟐 or^ - 2.

Since 𝒑 is negative therefore, the parabola opens to the left.

Step 4: Locate the focus by counting 𝑝 number of units from the vertex.

To locate the focus, count 2.5 units going left because 𝑝 = −2.5 and the parabola has horizontal principal axis.

The focus is 𝑭(−𝟏. 𝟓, 𝟑)

Step 5: Locate the directrix. (note: the distance of the focus and directrix from the vertex is equal but the direction is opposite)

The directrix is 𝒙 = 𝟑. 𝟓

Step 6 : Solve for the length of the Latus Rectum: 𝐿𝑅 = |4𝑝|

𝐿𝑅 = | 4 𝑝| 𝐿𝑅 = | 4 (− 2. 5 )| 𝐿𝑅 = |− 10 | 𝐿𝑅 = 1 0

Step 7: Find the endpoints of the Latus Rectum.

Clearly, the length of the latus rectum is 10 units. From the focus, count 5 units going up to locate the 1st^ endpoint 𝐴(− 1. 5 , 8 ), count another 5 units going down from the focus to locate the 2nd^ endpoint 𝐵(− 1. 5 , − 2 ). The distance from 𝐴 𝑡𝑜 𝐵 is 10 units.

Step 2: Identify the principal axis.

Since the given standard equation belongs to CASE 1 therefore, it has a vertical principal axis. From here, you can already imagine the graph of the parabola; it’s either opens upward or downward.

Step 3: Solve for the directed distance (𝑝).

The parabola opens downward since 𝒑 is negative.

Step 4: Locate the focus by counting 𝑝 number of units from the vertex.

To locate the focus count 2 units downward because 𝑝 = −2 and the parabola has vertical principal axis.

.

The focus is 𝑭(𝟎, −𝟐)

Step 5: Locate the directrix. (note: the distance of the focus and directrix from the vertex is equal but the direction is opposite)

To locate the directrix, count 2 units going up.

The directrix is 𝒚 = 𝟐

Step 6: Solve for the length of the Latus Rectum: 𝐿𝑅 = |4𝑝|

𝐿𝑅 = | 4 𝑝|

𝐿𝑅 = | 4 (− 2 )|

𝐿𝑅 = |− 8 |

𝐿𝑅 = 8

Step 7: Find the endpoints of the Latus Rectum.

Clearly, the length of the latus rectum is 8 units. From the focus count 4 units to the right to locate the 1st^ endpoint 𝐴(4, −2), count another 4 units to the left from the focus to locate the 2nd^ endpoint 𝐵(−4, −2). The distance from 𝐴 𝑡𝑜 𝐵 is 8 units.

The movement from the vertex to the focus is going down which means this

parabola has vertical principal axis.

Recall: Standard Equation of a Parabola that has a Vertical Principal Axis

𝑥^2 = 4 𝑝𝑦 Vertex at the origin (0,0)

Step 3: Identify the directed distance.

𝑝 is the directed distance from the vertex to the focus. Since the focus is 7 units below the vertex therefore, 𝑝 = −

Step 4: Substitute (ℎ, 𝑘)and 𝑝 to the chosen standard equation and simplify the equation.

𝑥^2 = 4 𝑝𝑦 Substitute only 𝑝 since the vertex is ( 0 , 0 )

𝑥^2 = 4 (− 7 )𝑦 Simplify

𝒙𝟐^ = −𝟐𝟖𝒚

Example 2: Write the standard equation of the parabola with the given properties: F (-4,,0) and directrix x=

Step 1 : Look for the vertex.

The vertex is not given so, graph the given data.

Since the vertex is halfway between the focus and the directrix, the vertex is at (−𝟏, 𝟎) , 𝒉 = −𝟏 and 𝒌 = 𝟎.

Step 2: Choose the correct standard equation to be used by determining the principal axis. Since the focus is located at the left side of the vertex, we have a horizontal principal axis.

Recall: Standard Equation of a Parabola that has a Horizontal Principal Axis

(𝑦 − 𝑘)^2 = 4 𝑝(𝑥 − ℎ) (^) Vertex at (ℎ, 𝑘)

Step 3: Identify the directed distance.

𝑝 is the directed distance from the vertex to the focus, we have 𝑝 = −

Step 4: Substitute (ℎ, 𝑘)and 𝑝 to the chosen standard equation and simplify the equation. (𝑦 − 𝑘)^2 = 4 𝑝(𝑥 − ℎ)

(𝑦 − 0 )^2 = 4 (− 3 )[𝑥 − (− 1 )]

𝒚𝟐^ = −𝟏𝟐(𝒙 + 𝟏)