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Parabola

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Parabola

1. 1. Geometric Definition of a Parabola: The collection of all the points P(x,y) in a a plane that are the samedistance from a fixed point, the focus, as they are from a fixed line called the directrix. y Focus (0,p) P x Verte x (h,k)
2. 2. y As you can plainly see the distance from the focus to the vertex is a and is the same distance from the vertex to the directrix! Neato! Focus (0,p) p 2p x Verte x pDirectrix y = -p (h,k) And the equation x = 4 py 2 is…
3. 3. y Directrix = a y Verte p x x (h,k) p Focus (0,-p)And theequation x = −4 py 2 is…
4. 4. y 2pDirectrix = −a x p p Verte Focus x (p,0) x (h,k)And the y = 4 px 2equation
5. 5. y p p Vertex Directrix = a x Focus (- (h,k) p,0) xAnd the y = −4 px 2equation
6. 6. STANDARD FORMS Vertex at (h, k ) 1) ( x − h) 2 = 4 p ( y − k ) Opens up Vertex at (h, k ) 2) ( x − h) 2 = −4 p ( y − k ) Opens down Vertex at (h, k ) 3) ( y − k ) 2 = 4 p ( x − h) Opens right Vertex at (h, k ) 4) ( y − k ) = −4 p ( x − h) 2 Opens leftI like to call standard form “Good Graphing Form”
7. 7. Graphing an Equation of a ParabolaStandard Equation of a Parabola (Vertex at Origin) x = 4 py 2 x = 12 y 2 ( 0, p ) focus ( 0, 3) y = −p directrix y = −3
8. 8. Graphing an Equation of a ParabolaStandard Equation of a Parabola (Vertex at Origin) y = 4 px 2 y = 12 x 2 ( p, 0 ) focus ( 3, 0 ) x = −p directrix x = −3
9. 9. Graphing an Equation of a ParabolaGraph the equation. Identify the focus and directrix of theparabola. 1. x = 2 y 2 1 4p = 2 p = 2  1 focus:  0,   2 1 directrix: y = − 2
10. 10. Graphing an Equation of a ParabolaGraph the equation. Identify the focus and directrix of theparabola. 2. y = 16 x 2 4 p = 16 p = 4 focus: ( 4, 0 ) directrix: x = −4
11. 11. Graphing an Equation of a ParabolaGraph the equation. Identify the focus and directrix of theparabola. 13. x = − y 2 1 4 14p = − p = − 4 16  1 focus:  0, −   16  1directrix: y = 16
12. 12. Graphing an Equation of a ParabolaGraph the equation. Identify the focus and directrix of theparabola. 4. y = −4 x 2 4 p = −4 p = −1 focus: ( − 1, 0 ) directrix: x = 1
13. 13. Writing an Equation of a ParabolaWrite the standard form of the equation of the parabolawith the given focus and vertex at (0, 0). 5. 0, 1( ) x = 4 py 2 x = 4( 1 ) y 2 x = 4y 2  1  6.  − , 0  y = 4 px 2  2  y 2 =4− 1 2 ( )x y = −2 x 2
14. 14. Writing an Equation of a ParabolaWrite the standard form of the equation of the parabolawith the given focus and vertex at (0, 0). 7. ( − 2, 0 ) y = 4 px 2 y = 4( − 2) x y = −8 x 2 2  1 8.  0,  x = 4 py 2  4 1 x = 4( 2 4 )y x =y 2
15. 15. Modeling a Parabolic Reflector9. A searchlight reflector is designed sothat a cross section through its axis is aparabola and the light source is at thefocus. Find the focus if the reflector is 3feet across at the opening and 1 foot deep. (1x5) = 4 p (1) . 2 2 y 2.25 = 4 p (1.5, 1) p = 2.25 = 9 225 4 400 16
16. 16. Notes Over 10.2Modeling a Parabolic Reflector10. One of the largest radio telescopes hasa diameter of 250 feet and a focal lengthof 50 feet. If the cross section of the radiotelescope is a parabola, find the depth. x = 4 py 2 250 = 125 x = 4( 50 ) y 2 2 x = 200 y 15,625 = 200 y 2 125 = 200 y 2 y = 78.1 ft
17. 17. General Form of any ParabolaAx + By + Cx + Dy + E = 0 2 2 *Where either A or B is zero! * You will use the “Completing the Square” method to go from the General Form to Standard Form,
18. 18. Graphing a Parabola: Use completing the square to convert a general form equation to standard conic form General formy2 - 10x + 6y - 11 = 0y2 + 6y + 9 = 10x + 11 + _____ 9 (y + 3)2 = 10x + 20 (y + 3)2 = 10(x + 2) Standard form aka: Graphing form (y-k)2 = 4p(x-h)