In this page focus question 4 we are going to find out the focus, vertex, equation of directrix and length of the latus rectum of the equation
y² +8x+2y+17=0.
Here the equation is in the standard form (y-k)²=4a(x-h).The following table gives the necessary details of the standard and vertex form of parabola.
Standard form |
Vertex form |
y² =4ax If a is positive, then it opens in the right hand side. If a is negative, then it opens in the left hand side. The focus is (a,0). The vertex is the origin (0,0) The equation of the directrix is x =-a The length of the latus rectum is 4a. |
(y-k)²=4a(x-h) If a is positive, then it opens in the right hand side. If a is negative, then it opens in the left hand side. The focus is (h+a, k) The vertex is (h,k) The equation of the directrix is x-h = -a The length of the latus rectum is 4a. |
Solution:
Here the equation y² +8x+2y+17=0 is in the quadratic equation form. Let us bring to the vertex form of equation.
y² +8x+2y+17=0.
y²+2y = -8x-17
y²+2y+1 = -8x-17+1(adding '1' on both sides)
(y+1) ² = -8x-16
(y+1) ² = -8(x+2)
This is of the form (y-k)²=4a(x-h) whose vertex is (h,k)
Here h=-2 and k=-1
and 4a = -8. So a = -8/4 =-2. Since a is negative the parabola opens up in the left hand side.
The focus is (h+a, k) = (-2-2,-1) = (-4,-1)
The vertex is (h,k) = (-2,-1)
The equation of the directrix is x+2 = +2
x=0
The length of the latus rectum is 4a =8
Parents and teachers help the students to solve the problem in the above method in focus question 4 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: