EQUATION OF A PARABOLA WITH VERTEX AND FOCUS

Example 1 :

Find the equation of the parabola, if the vertex is (4, 1) and the focus is (4, -3).

Solution :

From the given information the parabola is symmetric about y -axis and it opens down.

Distance between vertex and focus = a.

a = VF

= √[(4 - 4)² + (1 + 3)²]

= √[0 + 4²]

= √16

a = 4

Equation of the parabola :

(x - h)² = -4a(y - k)

Here, vertex (h, k) = (4, 1) and a = 4.

(x - 4)² = -4(4)(y - 1)

(x - 4)² = -16(y - 1)

Example 2 :

Find the equation of the parabola if the vertex is (0, 0) and the focus is (0, 4).

Solution :

From the given information the parabola is symmetric about y -axis and it opens up.

Distance between vertex and focus = a.

a = VF

= √[(0 - 0)² + (0 - 4)²]

= √(0 + 4²)

= √16

a = 4

Equation of the parabola :

(x - h)² = 4a(y - k)

Here, vertex (h, k) = (0, 0) and a = 4.

(x - 0)² = 4(4)(y - 0)

x² = 16y

Example 3 :

Find the equation of the parabola if the vertex is (1, 4) and the focus is (-2, 4).

Solution :

From the given information the parabola is symmetric about x -axis and it opens to the left.

Distance between vertex and focus = a

a = VF

= √[(1 + 2)² + (4 - 4)²]

  =  √(3² + 0)

 =  √9

a = 3

Equation of the parabola :

(y - k)² = -4a(x - h)

Here, vertex (h, k) = (1, 4) and a = 3.

(y - 4)² = -4(3)(x - 1)

(y - 4)² = -12(x - 1)

Example 4 :

Find the equation of the parabola if the vertex is (0, 0) and the focus is (5, 0).

Solution :

From the given information the parabola is symmetric about x -axis and it opens to the right.

Distance between vertex and focus = a

a = VF

= √[(0 - 5)² + (0 - 0)²]

  =  √(5² + 0)

 =  √25

a = 5

Equation of the parabola :

(y - k)² = 4a(x - h)

Here, vertex (h, k) = (0, 0) and a = 5.

(y - 0)² = 4(5)(x - 0)

y² = 20x

Example 5 :

Find the equation of the parabola with focus (-√2, 0) and directrix x = √2.

Solution :

From the given information the parabola is symmetric about x -axis and it opens left.

Distance between vertex and focus = distance between vertex and directrix.

a = √2

Equation of parabola :

(y - k)² = -4a(x - h)

(y - 0)² = -4√2(x - 0)

y² = -4√2x

Example 6 :

Find the equation of the parabola whose vertex is (5, -2) and focus (2, -2).

Solution :

From the given information the parabola is symmetric about x -axis and it opend left.

Distance between vertex and focus

= √(x₂ - x₁)² + (y₂ - y₁)²

a = √(2 - 5)² + (-2 + 2)²

a = √(-3)² + 0²

a = √9

a = 3

Equation of parabola :

(y - k)² = -4a(x - h)

(x - (-2))² = -4(3) (y - 5)

(x + 2)² = -12 (y - 5)

Expanding it using algebraic identities, we get

x² + 2(x)2 + 2² = -12 (y - 5)

x² + 4x + 4 = -12y + 60

x² + 4x + 12y + 4 - 60 = 0

x² + 4x + 12y - 56 = 0

Example 7 :

Find the equation of the parabola whose vertex (-1, -2) axis parallel to y-axis and passing through (3, 6).

Solution :

Since the axis is parallel to y-axis the required equation of parabola is 

(x - h)² = 4a(y - k)

(x - (-1))² = 4a(y - (-2))

(x + 1)² = 4a(y + 2)

The parabola passes through the point (3, 6)

(3 + 1)² = 4a(6 + 2)

4² = 4a(8)

16 = 32a

a = 16/32

a = 1/2

Applying the value of a, we get

(x + 1)² = 4(1/2)(y + 2)

(x + 1)² = 2(y + 2)

Expanding using algebraic identities, we get

x² + 2x(1) + 1² = 2y + 4

x² + 2x + 1 - 2y - 4 = 0

x² + 2x - 2y - 3 = 0

Example 8 :

Write an equation of the parabola with vertical axis of symmetry, vertex at the point (5, 1) and passing through the point (1, 3).

Solution :

The parabola is symmetric about y-axis and it is open upward.

(x - h)² = 4a(y - k)

Applying the given vertex (5, 1), we get

(x - 5)² = 4a(y - 1)

Since it passes through the point (1, 3), we get

(1 - 5)² = 4a(3 - 1)

(-4)² = 4a(2)

16 = 8a

a = 16/8

a = 2

Applying the value of a, we get

(x - 5)² = 4(2) (y - 1)

x² - 2x(5) + 5² = 8 (y - 1)

x² - 10x + 25 = 8 y - 8

x² - 10x - 8y + 25 + 8 = 0

x² - 10x - 8y + 33 = 0

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