FINDING THE VERTEX FOCUS AND DRECTRIX OF A PARABOLA FROM ITS EQUATION

In this section, you will learn how to find vertex, focus, equation of directrix and length of latus rectum of the parabola.

Before seeing example problems, let us remember some basic concepts about parabola. 

Parabola symmetric about x-axis and open right ward :

Standard form of parabola

y2  =  4ax

Parabola symmetric about x-axis and open left ward :

Standard form of parabola

y2  =  -4ax

Parabola symmetric about y-axis and open up ward :

Standard form of parabola

x2  =  4ay

Parabola symmetric about y-axis and open down ward :

Standard form of parabola

x2  =  -4ay

Now let us see some examples based on the above concept.

Example 1 :

Find the focus, vertex, equation of directrix and length of the latus rectum of the parabola

x2  =  5y

Solution :

From the given equation, the parabola is symmetric about y - axis and it is open upward.

x2  =  5y

4a  =  5

a  =  5/4

Vertex : V (0, 0)

Focus : F (0, 5/4)

Equation of directrix : y  =  -5/4

Length of latus rectum :  4a  =  4(5/4)  ==> 5

Example 2 :

Find the focus, vertex, equation of directrix and length of the latus rectum of the parabola

x2 - 8y - 2x + 17  =  0

Solution :

x2 - 8y - 2x + 17  =  0

x2 - 2x  =  8y - 17

x2 - 2x + 12 - 12  =  8y - 17

(x - 1)2  =  8y - 17 + 1

(x - 1)2  =  8y - 16

(x - 1)2  =  8(y - 2)

From the given equation, the parabola is symmetric about y - axis and it is open upward.

Let X  =  x - 1 and Y  =  y - 2

X2  =  8Y

4a  =  8

a  =  2

Referred to X and Y

X  =  x -  1 and Y  =  y - 2

Referred to x and y

x  =  X + 1 and y  =  Y + 2

Vertex (0, 0)

Focus (0, 2)

Equation of directrix

Y  =  -a

Y  =  -2

Length of latus rectum :

4a  =  4(2)  =  8

Vertex (1, 2)

Focus (1, 4)

Equation of directrix 

Y  =  0

Length of latus rectum :

4a  =  4(2)  =  8

Example 3 :

Find the focus, vertex, equation of directrix and length of the latus rectum of the parabola

 x2  =  -16y 

Solution :

From the given equation, the parabola is symmetric about y - axis and it is open downward.

x2  =  -16y

4a  =  16

a  =  4

Vertex : V (0, 0)

Focus : F (0, -4)

Equation of directrix : y  =  a  ==>  y  =  4

Length of latus rectum :  4a  =  4(4)  ==> 16

Example 4 :

Find the focus, vertex, equation of directrix and length of the latus rectum of the parabola

x2 + 4y - 6x + 17  =  0

Solution :

x2 + 4y - 6x + 17  =  0

x2 - 6x  =  -4y - 17

x2 - 6x + 32 - 32  =  -4y - 17

(x - 3)2  =  -4y - 17 + 9

(x - 3)2  =  -4y - 8

(x - 3)2  =  -4(y + 2) 

From the given equation, the parabola is symmetric about y - axis and it is open downward.

Let X  =  x - 3 and Y  =  y + 2

X2  =  -4Y

4a  =  4

a  =  1

Referred to X and Y

X = x -  3 and Y = y + 2

Referred to x and y

x  =  X + 3 and y  =  Y - 2

Vertex (0, 0)

Focus (0, -1)

Equation of directrix

Y  =  a

Y  =  1

Length of latus rectum :

4a  =  4(1)  =  4

Vertex (3, -2)

Focus (3, -3)

Equation of directrix 

Y  =  -1

Length of latus rectum :

=  4

Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Problems on Finding Derivative of a Function

    Mar 29, 24 12:11 AM

    Problems on Finding Derivative of a Function

    Read More

  2. How to Solve Age Problems with Ratio

    Mar 28, 24 02:01 AM

    How to Solve Age Problems with Ratio

    Read More

  3. AP Calculus BC Integration of Rational Functions by Partical Fractions

    Mar 26, 24 11:25 PM

    AP Calculus BC Integration of Rational Functions by Partical Fractions (Part - 1)

    Read More