Analytical Geometry Parabola Definition A parabola is the locus ( - PowerPoint PPT Presentation

Analytical Geometry

Parabola Definition A parabola is the locus ( راسم ) of a point P(x,y) moving in a plane such that: The distance from P(x,y) to the focus F = The distance from P(x,y) to the directrix D. D F

Standard Forms a Parabola  2 4 y ax (i) X- Parabola: V (0,0) y F a ( ,0) D   : directrix x a vertix  axis y : o axis focus 2 a V x a a F 2 a Latus Rectum length= 4 a directrix D '

Standard Forms a Parabola  2 4 x a y (i) Y- Parabola: V (0,0) F (0, ) a y   : directrix y a focus  axis x : o F 2 a 2 a a V x a vertix Latus Rectum length= 4 a D ' D axis directrix

All Standard Forms a Parabola  2 4 y ax   2 y 4 ax  2 x 4 ay   2 4 x ay

Example: Find the vertix, axis, focus, directrix, ends of L.R. and sketch the   2 x 8 y 0 Parabola Solution:   2   8 x y a 2 Vertix Focus Ends of Axis Directrix L.R V(0,0) F(0,-2) (4,-2) x=0 y=2 (-4,-2)

Parabola with Vertex at V(x 0 ,y 0 ) X- Parabola Y- Parabola             2  2    x x 4 a y y y y 4 a x x 0 0 0 0 General Equation     2 a x cx dy e 0 Y- Parabola     2 0 by cx dy e X- Parabola

Example: Find the vertix, axis, focus, directrix, ends of L.R. and     2 sketch the Parabola 2 y 16 x 4 y 30 0 Solution:     2 y 8 x 2 y 15 0     2 ( y 2 ) y 8 x 15 0      2 ( y 1) 1 8 x 15 0     2 ( y 1) 8 x 16       2 ( y 1) 8 x 2

      2 ( 1) 8 2 y x a = 2 Vertex Focus Ends of Axis Directrix L.R (2,1) (0,1) (0,5) y=1 x=4 (0,-3)

y directix y  1 F (0,1) axis V (2,1) X x x  4

Notes: * The distance between the focus and the directrix =2a. * The distance between the vertex and the directrix = the distance between the vertex and the focus = a. * The axis of symmetry and the directrix are perpendicular. * The axis of symmetry passes through the vertex and the focus. * To get the equation of parabola , we must know -- The type -- The vertex -- The value of a

Example: Find the equation of the parabola that has vertex V(-4,1) and has focus (-1,1). Solution V F -- The type X-parabola (+) -- The vertex V(-4,1) -- The value of a a=3      2   y 1 12 x 4

Example: Find the equation of the parabola that has focus (-1,1) and its equation of directrix is y=3. D Solution F -- The type y-parabola (-) -- The vertex V(-1,2) -- The value of a a=1         2 x 1 4 y 2