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

Finding the Vertex Focus and Directrix of a Parabola From its Equation :

In this section we can see 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 problems 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 YX  =  x -  1 and Y  =  y - 2 Referred to x and yx  =  X + 1 and y  =  Y + 2
 Vertex (0, 0)Focus (0, 2)Equation of directrixY  =  -aY  =  -2Length of latus rectum :4a  =  4(2)  =  8 Vertex (1, 2)Focus (1, 4)Equation of directrix Y  =  0Length 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 YX = x -  3 and Y = y + 2 Referred to x and yx  =  X + 3 and y  =  Y - 2
 Vertex (0, 0)Focus (0, -1)Equation of directrixY  =  aY  =  1Length of latus rectum :4a  =  4(1)  =  4 Vertex (3, -2)Focus (3, -3)Equation of directrix Y  =  -1Length of latus rectum :=  4

After having gone through the stuff given above, we hope that the students would have understood how to find the vertex, focus directrix and latus rectum of the parabola.

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