# FIND VERTEX FOCUS DIRECTRIX AND LATUS RECTUM OF PARABOLA

## About "Find Vertex Focus Directrix and Latus Rectum of Parabola"

Find Vertex Focus Directrix and Latus Rectum of Parabola :

Here we are going to see some example problems to understand the concept of finding vertex, focus, directrix, equation of latus rectum of the parabola.

## Details of Parabola with Vertex (0, 0)

Symmetric about x - axis and y - axis :

 Right V (0, 0)F (a, 0) LeftV (0, 0)F (-a, 0) UpV (0, 0)F (0, a) DownV (0, 0)F (0, -a)

Equation of latus rectum

 x = a x = -a y = a y = -a

Equation of directrix :

 x = -a x = a y = -a y = a

## Details of Parabola with Vertex (h, k)

Symmetric about x - axis and y - axis :

 Right V (h, k)F (h + a, k) LeftV (h, k)F (h-a, k) UpV (h, k)F (h, k + a) DownV (h, k)F (h, k-a)

Equation of latus rectum :

 x = h + a x = h - a y = k + a y = k - a

Equation of directrix :

 x = h-a x = h + a y = k - a y = k + a

## Find Vertex Focus Directrix and Latus Rectum of Parabola - Practice questions

Question 1 :

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

(i) y2  = 16x

Solution :

From the given information, we come to know that the given parabola is symmetric about x-axis and open right ward.

4a  =  16

a  =  4

Vertex : V (0, 0)

Focus : F (a, 0)  ==>  F(4, 0)

Equation of latus rectum : x = a ==>  x  =  4

Equation of directrix : x = -a ==>  x  =  -4

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

Let us look into the next problems on "Find Vertex Focus Directrix and Latus Rectum of Parabola"

(ii) x2 = 24y

Solution :

From the given information, we come to know that the given parabola is symmetric about y-axis and open upward.

4a  =  24

a  =  6

Vertex : V (0, 0)

Focus : F (0, a)  ==>  F(0, 6)

Equation of latus rectum : y = a ==>  y  =  6

Equation of directrix : y = -a ==>  y  =  -6

Length of latus rectum  =  4a  =  4(6)  =  24

(iii)  y2 = −8x

Solution :

From the given information, we know that the given parabola is symmetric about x-axis and left upward.

4a  =  8

a  =  2

Vertex : V (0, 0)

Focus : F (-a, 0)  ==>  F(-2, 0)

Equation of latus rectum : x = -a ==>  x  =  -2

Equation of directrix : x = a ==>  x  =  2

Length of latus rectum  =  4a  =  4(2)  =  8

(iv) x2 - 2x + 8y + 17  =  0

Solution :

x2 - 2x + 8y + 17  =  0

x2 - 2 ⋅ x  ⋅ 1 + 12 - 1+ 8y + 17  =  0

(x - 1)2 - 1 + 8y + 17  =  0

(x - 1)2  =  -8y - 16

(x - 1)2  =  -8(y + 2)

The parabola is symmetric about y-axis and open downward.

4a  =  8

a  =  2

Vertex : V(h, k)  ==>  (1, -2)

Focus : F (h, k - a) ==>  F (1, -4)

Equation of directrix : y = k + a ==>  y  =  -2 + 2  =  0

Length of latus rectum  =  4a  =  4(2)  =  8

(v) y2 - 4y - 8x + 12  =  0

Solution :

y2 - 4y - 8x + 12  =  0

y2 - 2 ⋅ y ⋅2 + 22 - 22 - 8x + 12  =  0

(y - 2)2 - 4  =  8x - 12

(y - 2)2  =  8x - 8

(y - 2)2  =  8(x - 1)

The parabola is symmetric about x-axis and open rightward.

4a  =  8

a  =  2

Vertex : V(h, k)  ==>  (1, 2)

Focus : F (h+a, k) ==>  F (3, 2)

Equation of directrix : x = h - a ==>  x  =  -1

Length of latus rectum  =  4a  =  4(2)  =  8

After having gone through the stuff given above, we hope that the students would have understood, "Find Vertex Focus Directrix and Latus Rectum of Parabola".

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