CONIC SECTIONS 1.3 Parabola

(d) Find the equation of the parabola with vertex (h, k) and focus (h+p, k) or (h, k+p). (e) Determine the vertex and focus of a parabola by completing the square (f) Introduce the applications of parabola such as suspension bridge, arch and reflector (a)Define a parabola. (b)Determine equation of a parabola with vertex (0,0) and focus (0,p). (c)Determine equation of a parabola with vertex (0,0) and focus (p,0).

N directrix x = – p P (x, y) F(p, 0) x Focus F(p,0) Directrix x = - p where p is a positive number. Vertex Let P(x,y) be any point on the parabola At the origin V(0,0) V(0, 0) y A parabola is a set of points that are equidistant from a fixed point (focus) and from a fixed line (directrix) Where, PF = PN

Equation of a parabola that has focus at (p,0) and directrix x = -p y x F(p,0) x = -p P(x,y) N

Vertex - origin (0,0) Symmetry to (a) x-axis : y 2 = 4px (b) y-axis : x 2 = 4py

Opens to right (a) Symmetry to x-axis : y 2 = 4px Opens to left x = - p F (p,0)x y V (0,0) x = - p F (p,0)x y V (0,0)

(b) Symmetry to y-axis : x 2 = 4py Opens upOpens down y = - p F (0,p) y xV (0,0) y = - p F (0,p) y x V (0,0)

EXAMPLE 1 Find the focus, directrix, vertex and axis of parabola for each of the following parabolas, and sketch its graph. (a) y 2 = 16x (b) x 2 = - 8y

Solution (a) y 2 = 16x Type : y 2 = 4px 4p = 16 p = 4 x = - 4 F (4,0)x y 0 Vertex : (0,0) Focus : (4,0) Directrix : x = - 4 Axis of parabola : x-axis

Solution Type : x 2 = 4py 4p = - 8 p = - 2 Vertex : (0,0) Focus : (0,-2) Directrix : y = 2 Axis of parabola : y-axis (b) x 2 = - 8y y = 2 F (0,-2) y 0x

(a) ( y – k ) 2 = 4p( x – h ) x = h - p F(h+p,k) x y 0 x = h - p F (h + p,k) y 0 x

(b) ( x – h ) 2 = 4p( y – k ) y = k - p F(h,k+p) x y 0 y = k - p F (h,k+p) y0 x

(y – k) 2 = 4p(x – h) or (x – h) 2 = 4p(y – k) standard form Ay 2 + Dx +Ey + F = 0 or Cx 2 + Dx +Ey + F = 0 general form REMARK

EXAMPLE 2 Write down the equation of parabola y 2 – 4y + 4x + 4 = 0 in standard form. Find the vertex, focus and directrix. Hence, sketch each graph.

Solution y 2 - 4y + 4x + 4 = 0 y 2 - 4y = - 4x - 4 ( y - 2 ) = - 4x - 4 ( y - 2 ) 2 = - 4x ( y – k ) 2 = 4p( x – h ) ( y - 2 ) 2 = 4(-1)( x - 0 ) F (-1,2) y 0 V (0,2) p = - 1 F ( -1, 2 ) x Directrix : x=1

EXAMPLE 3 Find an equation of the parabola described and sketch its graph. (a) Focus ( 0, -1 ); directrix the line y = 1 (b) Vertex ( -1, -1 ) ; axis of parabola is parallel to x-axis; containing the point (-3, 3).

Solution y 0 V (0,0) x 1 1 (x – h) 2 = 4p(y – k ) ( x – 0 ) 2 = 4p( y – 0) x 2 = 4py x 2 = 4(-1)y x 2 = -4y x 2 + 4y = 0 y = 1 F (0, -1) (a) Focus ( 0, -1 ); directrix y = 1

Solution F (-3,-1) y 0 V (-1,-1) x 2 2 ( y – k ) 2 = 4p( x – h ) ( y - (-1) ) 2 = 4p( x - (-1)) ( y + 1 ) 2 = 4p( x + 1 ) ( y + 1 ) 2 = 4(-2)( x + 1 ) y 2 + 2y + 8x + 9 = 0 x = 1 (b) Vertex (-1,-1); axis of parabola is parallel to x-axis; containing the point (-3,3). (-3,3) ( ) 2 = 4p(-3 + 1) 16 = -8p -2 = p ( y + 1 ) 2 = -8( x + 1 )

EXAMPLE 4 The cables of a suspension bridge are in the shape of a parabola, as shown in the figure. The towers supporting the cable are 600 feet apart and 80 feet high. If the cables touch the road surface midway between the towers, what is the height of the cable at a point 150 feet from the center of the bridge?

Solution 150 ft300 ft 80 ft x y (x – h) 2 = 4p(y – k ) x 2 = 4py = 320p = 4p(80) p = x 2 = 4( )y x 2 = 1125y (150,y) (300,80) V(0,0) (150) 2 = 1125 y y = 20 ft When x = 150, y = ?

Distance PF = Distance PN (a) Definition (b) Types of parabola’s equations F F P F N (i) y 2 = 4px (ii) x 2 = 4py Opens upOpens downOpens to rightOpens to left

(a) ( y – k ) 2 = 4p( x – h ) (b) ( x – h ) 2 = 4p( y – k )