FIND THE VERTEX OF THE PARABOLA
About "Find the vertex of the parabola"
Find the vertex of the parabola :
By comparing the given with any one of the following standard form, we can easily find the vertex of the parabola.
|
Equation |
Axis of symmetry |
Which side open |
|
(y-k)² = 4a(x-h) (y-k)² = -4a(x-h) (x-h)² = 4a(y-k) (x-h)² = -4a(y-k) |
x-axis x-axis y-axis y-axis |
Right ward Left ward Upward Downward |
In the above equations (h, k) is the vertex of the parabola.
Let us look in to some example problems based on the above concept.
Example 1 :
Find the vertex of the following parabola
x2 = - 4y
Solution :
x2 = - 4y
We can compare the above equation with the general form (x - h)2 = -4 a (y - k)
(x - 0)2 = - 4(y - 0)
Hence, the required vertex V(h, k) is (0, 0).
Example 2 :
Find the vertex of the following parabola
x2 − 2x + 8y + 17 = 0
Solution :
x2 − 2x + 8y + 17 = 0
Subtract 8y and 17 on both sides
x2 − 2x + 8y + 17 - 8y - 17 = -8y - 17
x2 − 2x = -8y - 17
Split the coefficient of x as the multiple of 2.
x2 − 2 ⋅x ⋅ 1 + 12 - 12 = -8y - 17
(x - 1)2 - 1 = -8y - 17
Add 1 on both sides
(x - 1)2 - 1 + 1 = -8y - 17 + 1
(x - 1)2 = -8y - 16
(x - 1)2 = -8(y + 2)
(x - 1)2 = -8(y - (-2))
The above equation exactly matches with the equation
(x - h)2 = -4a(y - k)
(h, k) ==> (1, -2)
Hence, the required vertex of the parabola is (1, -2).
Example 3 :
Find the vertex of the following parabola
y2 − 8x + 6y + 9 = 0
Solution :
y2 + 6y − 8x + 9 = 0
Add 8x and subtract 9 on both sides
y2 + 6y − 8x + 9 + 8x - 9 = 0 + 8x - 9
y2 + 6y = 8x - 9
Split the coefficient of y as the multiple of 2.
y2 + 2⋅ y ⋅ 3 + 32 - 32 = 8x - 9
(y - 3)2 - 9 = 8x - 9
Add 9 on both sides
(y - 3)2 - 9 + 9 = 8x - 9 + 9
(y - 3)2 = 8x
(y - k)2 = 4a (x - h)
(y - (-3))2 = 8(x - 0)
(h, k) ==> (0, -3)
Hence, the required vertex of the parabola is (0, -3).
Example 4 :
Find the vertex of the following parabola
x2 − 6x − 12y − 3 = 0
Solution :
x2 − 6x − 12y − 3 = 0
Add 12y and 3 on both sides
x2 − 6x − 12y − 3 + 12y + 3 = 0 + 12y + 3
x2 − 6x = 12y + 3
Split the coefficient of x as the multiple of 2.
x2 − 2⋅x⋅3 + 32 - 32 = 12y + 3
(x - 3)2 - 32 = 12y + 3
(x - 3)2 - 9 = 12y + 3
Add 9 on both sides
(x - 3)2 - 9 + 9 = 12y + 3 + 9
(x - 3)2 = 12y + 12
(x - 3)2 = 12(y + 1)
(x - 3)2 = 12(y - (-1))
(x - h)2 = 4a (y - k)
(x - 3)2 = 12(y - (-1))
(h, k) ==> (3, -1)
Hence, the required vertex of the parabola is (3, -1).
After having gone through the stuff given above, we hope that the students would have understood "Find the vertex of the parabola".
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