Examples on Equation of a Parabola

Example1: Find the length of the latus rectum, focus, and vertex, if the equation of the parabola is y² = 12x.

Solution:

Given, 

Equation of the parabola is y² = 12x

By comparing the given equation with the standard form y² = 4ax

4a = 12

⇒ a = 12/4 = 3

We know that,

Latus rectum of a parabola = 4a = 4 (3) = 12

Now, focus of the parabola = (a, 0) = (3, 0)

Vertex of the given parabola = (0, 0)

Example 2: Find the equation of the parabola which is symmetric about the X-axis, and passes through the point (-4, 5).

Solution:

Given,

Parabola is symmetric about the X-axis and has its vertex at the origin.

Thus, the equation can be of the form y² = 4ax or y² = -4ax, where the sign depends on whether the parabola opens towards the left side or right side.

Parabola must open left since it passes through (-4, 5) which lies in the second quadrant.

So, the equation will be: y² = -4ax

Substituting (-4, 5) in the above equation,

⇒ (5)² = -4a(-4)

⇒ 25 = 16a

⇒ a = 25/16

Therefore, the equation of the parabola is: y² = -4(25/16)x (or) 4y² = -25x.

Example 3: Find the coordinates of the focus, the axis, the equation of the directrix, and the latus rectum of the parabola x² = 16y.

Solution:

Given,

Equation of the parabola is: x² = 16y

By comparing the given equation with the standard form x² = 4ay,

4a = 16 ⇒  a = 4

Coefficient of y is positive so the parabola opens upwards.

Also, the axis of symmetry is along the positive Y-axis.

Hence,

Focus of the parabola is (a, 0) = (4, 0).

Equation of the directrix is y = -a, i.e. y = -4 or y + 4 = 0.

Length of the latus rectum = 4a = 4(4) = 16.

Example 4: Find the length of the latus rectum, focus, and vertex if the equation of a parabola is  2(x-2)² + 16 = y. 

Solution:

Given, 

Equation of a parabola is 2(x-2)² + 16 = y

By comparing the given equation with the general equation of a parabola y = a(x – h)² + k, we get

a = 2

(h, k) = (2, 16)

We know that, 

Length of latus rectum of a parabola = 4a

= 4(2) = 8

Now, focus= (a, 0) = (2, 0)

Now, Vertex = (2, 16)

Example 5: Equation of a parabola is x² – 12x + 4y – 24 = 0, then find its vertex, focus, and directrix.

Solution:

Given, 

Equation of the parabola is x² – 12x + 4y – 24 = 0

⇒ x² – 12x + 36 – 36 + 4y – 24 = 0

⇒ (x – 6)² + 4y – 60 = 0

⇒ (x – 6)² = -4(y + 15)

Obtained equation is in the form of  (x – h)² = -4a(y – k) 

-4a = -4 ⇒ a = 1

So, the vertex = (h, k) = (6, – 15)

Focus = (h, k – a) = (6, -15-1) = (6, -16)

Equation of the directrix is y = k + a

⇒ y = -15 + 1 ⇒ y = -14

⇒ y + 14 = 0

Standard form of a parabola is y = ax² + bx + c where a, b, and c are real numbers and a is not equal to zero. A parabola is defined as the set of all points in a plane that are equidistant from a fixed line and a fixed point in the plane.

In this article, we will understand what is a Parabola, the standard equation of a Parabola, related examples and others in detail.