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.
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Problem for practice: