Focus question 8

In this page focus question 8 we are going to find out the focus, vertex, equation of directrix and length of the latus rectum of the equation

x² +4y-6x+17=0.

Here the equation is in the standard form (x-h)²=4a(y-k).The following table gives the necessary details of the standard and vertex form of parabola.

 Standard form Vertex form

 x² =4ay  If a is positive, then it opens up. If a is negative, then it opens down. The focus is (0,a).  The vertex is the origin (0,0) The equation of the directrix is   y =-a The length of the latus rectum is   4a. (y-k)²=4a(x-h)  If a is positive, then it opens up. If a is negative, then it opens down. The focus is (h, k+a) The vertex is (h,k) The equation of the directrix is        y-k = -a The length of the latus rectum is 4a.

Solution:

Here the equation x² +4y-6x+17=0. is in the quadratic equation form. Let us bring to the vertex form of equation.

x² +4y-6x+17=0.

x²-6x = -4y-17

x²-6x+9 = -4y-17+9(adding '1' on both sides)

(x-3) ² =  - 4y-8

(x-3) ² =   -4(y+2)

This is of the form (x-h)²=4a(y-k) whose vertex is (h,k)

Here h=3 and k=-2

and 4a = -4. So a = -4/4 =-1. Since a is negative, it opens down.

The focus is (h, k+a)  =  (3,-2-1) = (3,-3)

The vertex is (h,k)                    = (3,-2)

The equation of the directrix is y-k = -a

y-(-2)= -1

y=-3

The length of the latus rectum is 4a = 4

Parents and teachers help the students to solve the problem in the above method in focus question 8 and they can guide them to solve the following problem using the above method.

The other three standard forms  and vertex forms of parabola are discussed in the focus worksheet.

If you have any doubt you can contact us through mail, we will help you to solve the problem.

Problem for practice:

1.         Find the focus, vertex, equation of directrix and length of latus rectum of the parabola x²+12y-2x+3=0
2.    Find the focus, vertex, equation of directrix and lenth of the latus rectum of the parabola x²+4y+2x-3=0 