Warm Up circle hyperbola circle Find the standard form of the equation by completing the square. Then identify and graph the conic. d.

Find the standard form of the equation by completing the square Find the standard form of the equation by completing the square. Then identify and graph the conic. d.

Objective: To find equations of parabolas and to graph them.

Conic Sections - Parabolas The parabola has the characteristic shape shown above. A parabola is defined to be the “set of points the same distance from a point and a line”.

Conic Sections - Parabolas Focus Directrix The line is called the directrix and the point is called the focus.

Conic Sections - Parabolas Axis of Symmetry Focus Directrix Vertex The line perpendicular to the directrix passing through the focus is the axis of symmetry. The vertex is the point of intersection of the axis of symmetry with the parabola.

Conic Sections - Parabolas Focus d1 Directrix d2 The definition of the parabola is the set of points the same distance from the focus and directrix. Therefore, d1 = d2 for any point (x, y) on the parabola.

Circles

Hyperbolas

Finding the Focus and Directrix 6.5 Parabolas Finding the Focus and Directrix

Conic Sections - Parabolas Focus y = ax2 p Directrix p We know that a parabola has a basic equation y = ax2. The vertex is at (0, 0). The distance from the vertex to the focus and directrix is the same. Let’s call it p.

Conic Sections - Parabolas Focus ( ?, ?) y = ax2 p Directrix ??? ( 0, 0) p Find the point for the focus and the equation of the directrix if the vertex is at (0, 0).

Conic Sections - Parabolas Focus ( 0, p) y = ax2 p Directrix ??? ( 0, 0) p The focus is p units up from (0, 0), so the focus is at the point (0, p).

Conic Sections - Parabolas Focus ( 0, p) y = ax2 p Directrix ??? ( 0, 0) p The directrix is a horizontal line p units below the origin. Find the equation of the directrix.

Conic Sections - Parabolas Focus ( 0, p) y = ax2 p Directrix y = -p ( 0, 0) p The directrix is a horizontal line p units below the origin or a horizontal line through the point (0, -p). The equation is y = -p.

Conic Sections - Parabolas ( x, y) Focus ( 0, p) d1 y = ax2 d2 Directrix y = -p ( 0, 0) The definition of the parabola indicates the distance d1 from any point (x, y) on the curve to the focus and the distance d2 from the point to the directrix must be equal.

Conic Sections - Parabolas ( x, ax2) Focus ( 0, p) d1 y = ax2 d2 Directrix y = -p ( 0, 0) However, the parabola is y = ax2. We can substitute for y in the point (x, y). The point on the curve is (x, ax2).

Conic Sections - Parabolas ( x, ax2) Focus ( 0, p) d1 y = ax2 d2 Directrix y = -p ( 0, 0) ( ?, ?) What is the coordinates of the point on the directrix immediately below the point (x, ax2)?

Conic Sections - Parabolas ( x, ax2) Focus ( 0, p) d1 y = ax2 d2 Directrix y = -p ( 0, 0) ( x, -p) The x value is the same as the point (x, ax2) and the y value is on the line y = -p, so the point must be (x, -p).

Conic Sections - Parabolas ( x, ax2) Focus ( 0, p) d1 y = ax2 d2 Directrix y = -p ( 0, 0) ( x, -p) d1 is the distance from (0, p) to (x, ax2). d2 is the distance from (x, ax2) to (x, -p) and d1 = d2. Use the distance formula to solve for p.

Conic Sections - Parabolas d1 is the distance from (0, p) to (x, ax2). d2 is the distance from (x, ax2) to (x, -p) and d1 = d2. Use the distance formula to solve for p. d1 = d2

Conic Sections - Parabolas Therefore, the distance p from the vertex to the focus and the vertex to the directrix is given by the formula 𝒑= 𝟏 𝟒𝒂 𝒂= 𝟏 𝟒𝒑 𝒐𝒓 − 𝟏 𝟒𝒑 depending on the direction of the parabola.

6.5 Parabolas Parabola Equation

Parabola Equation center at the origin (0,0) 𝒑= 𝟏 𝟒𝒂 𝒂= 𝟏 𝟒𝒑 𝒐𝒓 − 𝟏 𝟒𝒑 Parabola Equation center at the origin (0,0) 𝟏. 𝒚= 𝟏 𝟒𝒑 𝒙 𝟐 𝟐. 𝒚=− 𝟏 𝟒𝒑 𝒙 𝟐 𝟑. 𝒙= 𝟏 𝟒𝒑 𝒚 𝟐 𝟒. 𝒙=− 𝟏 𝟒𝒑 𝒚 𝟐 p p p p p (p, 0) p p)

Conic Sections - Parabolas Using transformations, we can shift the parabola y=ax2 horizontally and vertically. The vertex is shifted from (0, 0) to (h, k). Recall that when “a” is positive, the graph opens up. When “a” is negative, the graph reflects about the x-axis and opens down.

Parabola Equation center at (h,k) 𝒚= 𝟏 𝟒𝒑 𝒙−𝒉 𝟐 +𝒌 𝒚=− 𝟏 𝟒𝒑 𝒙−𝒉 𝟐 +𝒌 𝒙= 𝟏 𝟒𝒑 𝒚−𝒌 𝟐 +𝒉 𝒙=− 𝟏 𝟒𝒑 𝒚−𝒌 𝟐 +𝒉

Examples Graph a parabola. Find the vertex, focus and directrix. 6.5 Parabolas Examples Graph a parabola. Find the vertex, focus and directrix.

Example 1 Find the focus and directrix of each parabola 𝒑= 𝟏 𝟒𝒂 a. 𝑦=2 𝑥 2 b. x= 1 20 𝑦 2 𝑝= 1 4(2) 𝑝= 1 4( 1 20 ) p= 1 8 p=5 Focus: 0, 1 8 Directrix: 𝑦=− 1 8 Focus: 5,0 Directrix: x=−5

Example 2 Find an equation of the parabola with vertex (0,0) and directrix x = 2. Sketch the information. 𝑥=2 𝒙=− 𝟏 𝟒𝒑 𝒚 𝟐 𝒙=− 𝟏 𝟒(𝟐) 𝒚 𝟐 𝒙=− 𝟏 𝟖 𝒚 𝟐

Example 3 Graph the parabola. Find the vertex, focus, and directrix. The vertex is (-2, -3). The parabola opens up.

Find the focus and directrix. The focus and directrix are “p” units from the vertex where The focus and directrix are 2 units from the vertex.

Find the focus and directrix. 2 Units Focus: (-2, -1) Directrix: y = -5

Example 4 Write the equation of a parabola with vertex at (3, 2) and focus at (-1, 2). Plot the known points. What can be determined from these points? The parabola opens the the left and has a model of x = − 𝟏 𝟒𝒑 (y – k)2 + h.

Example 4 Write the equation of a parabola with vertex at (3, 2) and focus at (-1, 2). x = − 𝟏 𝟒𝒑 (y – k)2 + h. The distance from the vertex to the focus is 4, so p = 4. x = − 𝟏 𝟒𝒑 (y – 2)2 + 3 x = − 𝟏 𝟒(𝟒) (y – 2)2 + 3 x = − 𝟏 𝟏𝟔 (y – 2)2 + 3

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