Conics: Parabolas on a Coordinate Plane

On Tuesday, we learned how to make parabolas on a coordinate plane.

Recall what you learned about parabolas last year. Remember the standard rules of a parabolas shape where if |a| = 1, then the shape would be normal but if |a| > 1 then the parabola would get narrower and if |a| < 1 then the parabola would get wider.

Recall the standard form of parabolas.

(y-k) = a(x-h)^2 or
(x-h) = a(y-k)

For our Conics unit, we have learned the General Form for a Parabola:

Ax^2 + Cx + Dy + E = 0 or
By^2 + Cx + Dy + E = 0

We use completion of squares to convert to standard form to find the vertex.

Example 1: Find the vertex, direction and shape of parabola, axis of symmetry, domain and range for each conic.

2x^2 + 4x - 4y +22 = 0

We can't find any data with this kind of equation so ee need to complete the square in order to solve this equation.

2x^2 + 4x - 4y + 22 = 0
2(x^2 + 2x) - 4y = -22
2(x^2 + 2x + 1) - 4y = -22 + 2
(2(x + 1)^2 - 4y = -20) / 2
(x + 1)^2 = 2y = -10

Rearrange the equation...

-2y = -(x+1)^2 - 10

Now isolate the y...

(-2y = -(x+1)^2 - 10) / -2
y = (1/2)(x + 1)^2 + 5

Now we can fill in the missing data with this equation.

Vertex: (-1, 5)
Direction: Upwards
Shape: Wide
Axis of Symmetry: x = -1
Domain: (-∞, ∞)
Range: (-5, ∞)

Example 2: If the parabola y^2 + x - 6y - m = 0 passes through the point (-3,5) find the vertex.

First we need to solve for m. We do this by plugging in the values of x and y.

(5)^2 + (-3) - 6(5) - m = 0
25 - 3 - 30 = m
m = -8

Now we can solve the equation.

y^2 + x - 6y - (-8) = 0
y^2 - 6y + x + 8 = 0
(y^2 - 6y + 9) +x + 8 - 9 = 0
(y - 3)^2 + x - 1 = 0
(x-1) = -(y-3)^2

Therefore, the vertex is at (1,3).

QED

Also a reminder that all assignments are due in Wednesday. Please submit any work that you have by then!

Good luck! :D