FINDING EQUATION OF PARABOLA GIVEN FOCUS AND VERTEX

Problem 1 :

Find the equation of parabola with focus (5, 0) and vertex (5, 3).

Solution :

Always foucs and vertex will lie on the same line. The focus is below the vertex, the parabola opens down.

(x - h)² = -4a(y - k)

Here (h, k) is (5, 3)

(x - 5)² = -4a(y - 3) ----(1)

To find "a", let us find the distance between focus and vertex,

Distance between two points = √(x₂ - x₁)² + (y₂ - y₁)²

= √(5 - 5)² + (3 - 0)²

= √9

a = 3

By applying the value of a in (1), we get

(x - 5)² = -4(3)(y - 3)

(x - 5)² = -12(y - 3)

Problem 2 :

Find the equation of parabola whose focus is at (6, 3) and the vertex (2, 3) is given by.

Solution :

Always foucs and vertex will lie on the same line. The focus is at next to vertex. So the parabola opens right and it is symmetric about x-axis.

(y - k)² = 4a(x - h)

Here (h, k) is (2, 3)

(y - 3)² = 4a(x - 2) ----(1)

To find "a", let us find the distance between focus and vertex,

Distance between two points = √(x₂ - x₁)² + (y₂ - y₁)²

= √(6 - 2)² + (3 - 3)²

= √4² + 0²

a = 4

By applying the value of a in (1), we get

(y - 3)² = 4(4)(x - 2)

(y - 3)² = 16(x - 2)

Problem 3 :

Find the equation of parabola whose focus is at (-2, 4) and the vertex (1, 4) is given by.

Solution :

Always foucs and vertex will lie on the same line. The focus is in front of vertex. So the parabola opens left and it is symmetric about x-axis.

(y - k)² = -4a(x - h)

Here (h, k) is (1, 4)

(y - 4)² = -4a(x - 1) ----(1)

To find "a", let us find the distance between focus and vertex,

Distance between two points = √(x₂ - x₁)² + (y₂ - y₁)²

= √(1 - (-2))² + (4 - 4)²

= √3² + 0²

a = 3

By applying the value of a in (1), we get

(y - 4)² = -4(3)(x - 1)

(y - 4)² = -12(x - 1)