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

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