Examples of Parabolas

5 SOLVED PROBLEMS OF PARABOLA

EXAMPLE#1

Find the focus and the directrix of

The following parbola

y² = 16x

x = 16y²

4p = 16

4        4

p = 4

Focus = (h+4,0) Directrix: y=k – 4

Focus = (4,0) Directrix: x = -4

GRAPH:

EXAMPLE#2

Find the directrix,focus, and vertex of

The parabola

y = ½ x²

(x – h)² = 4p(y – k) Standard Form

1/2 x² = y

x² = 2y

2 = 4p

4    4

p = ½

(x – 0)² = 4(1/2)(y – 0)²

Vertex (0,0) Focus (0,1/2) Directrix: y = -1/2

GRAPH:

EXAMPLE#3

Determine the vertex,focus, and the

Directrix of the parbola from this

Equation and graph

(y – 1)² = 8(x + 5)

Vertex (h,k) = (-5,1)

Focus = 4p = 8

              4      4

p = 2 Focus: (-5+2,1)

Focus: (-3,1)

Directrix: x = - 5 - 2

Directrix: x = -7

GRAPH:

EXAMPLE#4

Determine the diractrix , focus and

The vertex of the parabola and graph

(x – 3)² = 3(y + 1)

Vertex (h,k) = (3,-1)

4p = 3

4      4

Focus (3,-1 + ¾) Focus (3, - ¼)

Directrix: y = 1 – ¾  Directrix: y = -7/4

GRAPH:

EXAMPLE#5

Find the Standard Form of a parabola

from the given directrix and vertex and

graph

Vertex (0,4) Directrix: y = 2

SOLUTION

P = 2

Focus = (0,4+2) Focus (0,6)

(x-h)² = 4p(y-4)

(x-0)² = 4(2)(y – 4)

x² = 8(y – 4)

x² = 8y – 32

y = 1/8 (x² + 32) S.F

GRAPH:

And that's all :)

Examples Of Circles

5 SOLVED PROBLEMS OF CIRCLE

EXAMPLE# 1

Find the center and radius of this circle:

x² + y² – 8x + 2y – 19 = 0

SOLUTION:

x² – 8x + 16 + y² + 2y + 1 = 19+16+1

(x – 4)² + (y + 1)² = ⎷36

(x – 4)² + (y + 1)² = 6²

Center: (4,-1) Radius: r = 6
GRAPH:

EXAMPLE#2

What is the center and radius of the

circle indicated by this equation?

(x – 2)² + (y)² = 36

SOLUTION:

(x – 2)² + (y)² = ⎷36

(x – 2)² + (y)² = 6²

Center: (2,0) Radius: r = 6

GRAPH:

EXAMPLE#3

What is the equation of the circle

with a center at (4,-5) and a point

in the circle at (4,-2).What is the

radius?

SOLUTION:

(x-h)² + (y-k)² = r² Standard Formula

(x-4)² + (y-(-5))² = r² 

(4-4)² + (-2+5)² = r²

(0)² + (3)² = r²

9 = r²

r = 3²

^GRAPH:

Center: (4,-5) Radius: r = 3

EXAMPLE#4

Determine the center and the radius

Of the circle from this equation and graph

x² + y² – 8x + 6y + 9 = 0

SOLUTION:

x² – 8x + 16 + y² + 6y + 9 = -9 + 9 + 16

(x-4)² + (y+3)² = 16

(x-4)² + (y+3)² = 16

(x-4)² + (y+3)² = 4²

Center: (4,-3) Radius: r = 4

GRAPH:

EXAMPLE#5

Determine the center and the radius

Of the circle from this equation and

graph

x² + y² – 6x + 6y – 18 = 0

x² – 6x + 9 + y² + 6y + 9 = 18 + 9 + 9

(x-3)² + (y+3)² = 36

(x-3)² + (y+3)² = 36

(x-3)² + (y+3)² = 6²

Center: (3,-3) Radius: r = 6

GRAPH:

1st Lesson Conic Sections

1st Lesson: Conic Sections

The conic sections are the nondegenerate curves generated by the intersections of a plane with one or two nappes of a cone. For a plane perpendicular to the axis of the cone, a circle is produced. For a plane that is not perpendicular to the axis and that intersects only a single nappe, the curve produced is either an ellipse or a parabola (Hilbert and Cohn-Vossen 1999, p. 8). The curve produced by a plane intersecting both nappes is a hyperbola (Hilbert and Cohn-Vossen 1999, pp. 8-9

GENERAL EQUATION
Ax²­+Bxy+Cy²+Dx+Ey+F=0


THE DISCRIMINANT FORMULA FOR CONIC SECTION

 b² – 4ac > 0 HYPERBOLA
b²  - 4ac = 0 PARABOLA
b²  - 4ac <  0 CIRCLE

         where A = C B = 0
b²  - 4ac <  0 ELLIPSE
         where A≠c B=0

EXAMPLES:
1. G.E
 2y² + 20x + 2y – 1 = 0
 A= 0  B=  0  C= 2
b² – 4ac = (0)⁰ – 4 (0)(2)
                = 0 – 0
                = 0 PARABOLA

S.F : (h,k) (y-k)² = 4a(x-h)

2y² + 20x + 12y – 1 = 0

2y² + 12y       20x – 1 = 0

2(y² + 6y + 9) = -20x + 1

 2(y+3)²  = -20x + 1 + 9

    2

(y+3)² = -10x + 10

S.F = (y+3)² = -10x + 10

h = 5 k = 3

 2. G.E

6x² – 5y + 36x + 20y – 16 = 0

A= 6  B= 0  C= -5

b² – 4ac = (0)² – 4 (6)(-5)

= 120 < 0  HYPERBOLA

S.F

6x² + 36x ____ -5y + 20y – 16 = 0

6(x²+6x+9) – 5(y² – 4y + 4)= 16 + 9 + 4

6(x+3)² – 5(y-2)² = 29

              30

(x +3)² – (y-2)²

     5            6

   10          10

(h = 3) (k= -2)