FIND THE FOCUS DIRECTRIX VERTEX AND AXIS OF SYMMETRY FOR THE PARABOLA

Parabola Opens right ward

parabola-open-right

Parabola Opens Left ward

parabola-open-left.png

Parabola Opens Up ward

parabola-open-up

Parabola Opens Down ward

parabola-open-down.png

Graph the parabola and identify the vertex, directrix, and focus.

Problem 1 :

y2 = -12x

Solution:

y2 = -12x

graph-the-parabola-q1

The above equation of parabola is in standard form. The parabola is symmetric about x-axis and it opens to the left.

4a = 12

a = 3

Vertex:

(0, 0)

Equation of directrix:

x = a

x = 3

Focus:

F(-a, 0) = F(-3, 0)

Problem 2 :

x2 = 8y

Solution:

x2 = 8y

graph-the-parabola-q2.png

The above equation of parabola is in standard form. The parabola is symmetric about y-axis and it opens to the up.

4a = 8

a = 2

Vertex:

(0, 0)

Equation of directrix:

y = -a

y = -2

Focus:

F(0, a) = F(0, 2)

Problem 3 :

6(x + 1)2 + 12(y - 3) = 0

Solution:

6(x + 1)2 + 12(y - 3) = 0

6(x + 1)2 = -12(y - 3)

(x + 1)2 = -12/6 (y - 3)

(x + 1)2 = -2(y - 3)

Let X = x + 1 and Y = y - 3

X2 = -2Y

The above equation of parabola is in standard form. The parabola is symmetric about y-axis and it opens to the down.

graph-the-parabola-q3.png

4a = 2

a = 1/2

Vertex :

(0, 0)

X = 0 and Y = 0

x + 1 = 0 and y - 3 = 0

x = -1 and y = 3

The vertex is (-1, 3).

Equation of directrix:

Y = a

Y = 1/2

Y = 0.5

y - 3 = 0.5

y = 0.5 + 3

y = 3.5

Focus:

F(0, -a) = F(0, -0.5)

X = 0 and Y = -0.5

x + 1 = 0 and y - 3 = -0.5

x = -1 and y = 2.5

The focus is (-1, 2.5).

Problem 4 :

y2 - 12(x + 2) = 0

Solution:

y2 - 12(x + 2) = 0

y2 = 12(x + 2)

Let X = x + 2

y2 = 12X

graph-the-parabola-q4.png

The above equation of parabola is in standard form. The parabola is symmetric about x-axis and it opens to the left.

4a = 12

a = 3

Vertex:

(0, 0)

X = 0 and Y = 0

x + 2 = 0 and y = 0

x = -2 and y = 0

The vertex is (-2, 0).

Equation of directrix:

X = -a

X = -3

x + 2 = -3

x = -5

Focus:

F(a, 0) = F(3, 0)

x + 2 = 3

x = 1

The focus is (1, 0).

Problem 5 :

(x + 2)2 = -8(y + 2)

Solution:

(x + 2)2 = -8(y + 2)

Let X = x + 2 and Y = y + 2

X2 = -8Y

graph-the-parabola-q5.png

The above equation of parabola is in standard form. The parabola is symmetric about y-axis and it opens to the down.

4a = 8

a = 2

Vertex:

(0, 0)

X = 0 and Y = 0

x + 2 = 0 and y + 2 = 0

x = -2 and y = -2

The vertex is (-2, -2).

Equation of directrix:

Y = a

Y = 2

y + 2 = 2

y  = 0

Focus:

F(0, -a) = F(0, -2)

X = 0 and Y = -2

x + 2 = 0 and y + 2 = -2

x = -2 and y = -4

The focus is (-2, -4).

Problem 6 :

(y - 1)2 = -8x

Solution:

(y - 1)2 = -8x

Let Y = y - 1

Y2 = -8x 

graph-the-parabola-q6.png

The above equation of parabola is in standard form. The parabola is symmetric about x-axis and it opens to the left.

4a = 8

a = 2

Vertex:

(0, 0)

Y = 0

y - 1 = 0

y = 1

The vertex is (0, 1).

Equation of directrix:

x = a

x = 2

Focus:

F(-a, 0) = F(-2, 0)

x = -2 and Y = 0

 y - 1 = 0

x = -2 and y = 1

The focus is (-2, 1).

Recent Articles

  1. Factoring Exponential Expression Using Algebraic Identities Worksheet

    Mar 14, 24 10:44 PM

    Factoring Exponential Expression Using Algebraic Identities Worksheet

    Read More

  2. Positive and Negative Numbers Connecting in Real Life Worksheet

    Mar 14, 24 10:12 AM

    Positive and Negative Numbers Connecting in Real Life Worksheet

    Read More

  3. Positive and Negative Numbers Connecting in Real Life

    Mar 14, 24 09:52 AM

    Positive and Negative Numbers Connecting in Real Life

    Read More