What is the focus of the parabola?
y=−14x2−2x−2
Enter your answer in the boxes.
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The equation of a parabola is y=18x2−4x+41 .
What is the equation of the directrix?
y = 14
y = 11
y = 7
y = 2
The equation of a parabola is 132(y−2)2=x−1 .
What are the coordinates of the focus?
(9, 2)
(−7, 2)
(1, 10)
(1, −6)
y = -14x^2 - 2x - 2
We want to get to the form 4p (y - k) = ( x - h)^2
Where (h, k) is the vertex and p is the distance from the vertex to the focus
We can factor this as
y = -14 [x^2 + (1/7)x + (1/7) ]
Complete the square inside the brackets
Take (1/2)of (1/7) = (1/14)....square this = 1/196 ...add it and subtract it
So we have
y = -14 [ x^2 +(1/7)x + 1/196 + (1/7) - (1/196) ]
Factor the first three terms and simplify the last two.....so we have
y = -14 [ (x + 1/14)^2 + 27/196] simplify
y = -14 ( x + 1/14)^2 - 27/14 add 27/14 to both sides
(y +27/14) = -14 ( x + 1/14)^2 multiply both sides by -1/14
(-1/14)(y + 27/14) = ( x + 1/14)^2
Since -1/14 = 4p....divide both sides by 4.....then p = - 1/56
So... the vertex is (h, k) = ( -1/14, -27/14 )
And the focus is given by ( h , k + p) =
( -1/14, - 27/14 - 1/56) =
(- 1/14, - 109/56 )
Here's the graph : https://www.desmos.com/calculator/moljoeobcv
The equation of a parabola is y=18x^2−4x+41 .
What is the equation of the directrix?
We want the form 4p(y - k) = (x - h)^2 where (h, k) is the vertex and p is the distance between the vertex and the directrix
Factor out 18
y = 18 ( x^2 - (2/9)x + 41/18 )
Complete the square inside the parentheses
Take (1/2) of (2/9) = (1/9) .....square it = (1/81)....add and subtract it
y = 18 [ x^2 - (2/9)x + 1/81 + 41/18 - 1/81 ]
factor the first three terms, simplify the last two
y = 18 [ (x - 1/9)^2 + 367/162] simplify
y = 18 ( x - 1/9)^2 + 367/9
subtract 367/9 from both sides
(y - 367/9) = 18 ( x - 1/9)^2 divide both sides by 18
(1/18) (y - 367/9) = ( x - 1/9)^2
Since 4p = 1/18, divide both sides by 4....so p = 1/72
The vertex is ( 1/9, 367/9)
And the equation of the directrix is y = 367/9 - p → y = 367/9 - 1/72 →
y = 2935 / 72
Here's the graph : https://www.desmos.com/calculator/qjwolpipur
The equation of a parabola is 132(y−2)^2 = x−1 .
What are the coordinates of the focus?
We want the form
4p ( x - h) = (y - k)^2 wher (h, k) is the vertex and p is the distance between the vertex and the focus
132 (y - 2)^2 = (x - 1) rearrange as
(x - 1) = 132 ( y - 2)^2 divide both sides by 132
(1/132) ( x - 1) = ( y - 2)^2
4p = (1/132) divide both sides by 4 and p = 1/528
And the focus is given by
( h + p , k) = ( 1 + 1/528 , 2) = ( 529/528, 2 )
Here's the graph : https://www.desmos.com/calculator/cqctvmc09d