PLZ!!!! can someone help me with my algbra 2 homework plz need help with parabola and circles PLZ!!!?

Question

Ex. (x-1)sq=12(y-1) Identify the coordinates of the vertex and focus, the equations of the axis of symmetry and directrix, and the direction of the opening of the parabola with the given equation. then find the length of the latus rectum and graph the parabola. PLZ SOMEONE WHO CAN DO THIS PLZ I SWAER THIS MEANSS SO MUCH PLZ

Answer 1

(x - 1)^2 = 12(y - 1)

Your first step would be to convert this into the form
y = a(x - h)^2 + k

Once you get it into this form, your vertex will be at (h, k)

First, let's expand the right hand side.

(x - 1)^2 = 12y - 12

Move -12 to the left hand side.

(x - 1)^2 + 12 = 12y

Now, divide everything by 12.

(1/12) (x - 1)^2 + 1 = y, or, switching it around,

y = (1/12) (x - 1)^2 + 1

To determine which direct the parabola opens, it is dependent on the value of "a". If "a" is positive, the parabola opens upward; if "a" is negative, it opens downward. In our case, a = (1/12), so the parabola opens upward.

Answer 2

Given the equation

12(y - 1) = (x - 1)²

determine its properties.

Of the two variables, x and y, in the equation, one is linear and the other is squared. This indicates that the geometric form is a parabola.

You need to know the basic properties of the parabola. If written in the form

4a(y - k) = (x - h)²

you can readily obtain much of the information you seek. Our equation is:

12(y - 1) = (x - 1)²

Note that the y term is linear and the x term is squared. This indicates a vertical parabola. The two variables are on opposite sides of the equal sign and have the same sign. This indicates that the parabola opens in the positive direction, in this case, up.

Now let's calculate the constants.

h = 1
k = 1
4a = 12
a = 3

We can see from the equation that (h,k) = (1,1). That is the vertex. Since the parabola is vertical so is the axis of symmetry. And it runs thru the vertex. So the axis of symmetry is x = 1.

The directrix is outside the parabola, in this case below it. And it is perpendicular to the axis of symmetry. The distance from the vertex to the directrix is a = 3.

So the equation of the directrix is:

y = k - a
y = 1 - 3 = -2

The focus is a distance a from the vertex in the opposite direction from the directrix. It is inside the parabola, in this case, above it. The focus is located at:

x = h = 1
Since the parabola is vertical, the x coordinate is the same as that of the vertex.

y = k + a = 1 + 3 = 4

So the focus is at the point (1,4).

The latus rectum is a line segement parallel to the directrix, that runs inside the parabola from one side of the parabola, thru the focus, to the other side of the parabola. Its length is 4a.

4a = 4*3 = 12

It runs from (1-2a,4) to (1+2a,4). In other words, from
(1 - 2*3,4) to (1 + 2*3,4). From (-5,4) to (7,4).

You'll have to graph it yourself. Plug in the vertex as one point and the endpoints of the latus rectum as two more. Then pick a couple points for x on either side of the vertex and calculate the associated y values.

Please study this so you'll know how to figure this stuff out on your own for your next parabola. Good luck.

Answer 3

You have a nearly perfect equation for answering your question.

(x-1)^2 = 12(y-1)

Vertex and line of symmetry first. When x=1 and y=1, you get a minimum (0=0). Your vertex is at (1,1). Easy. And since this parabola opens upward -- when x gets large (positive or negative), y gets large positive -- your line of symmetry is vertical, passing through the vertex. The equation that fits this is x=1 (line of symmetry).

Focus and latus rectum next. You know that the latus rectum is four times the focal length. You want the form (x-h)^2 = 4a(y-k). In your case, that's

(x-1)^2 = 4*3(y-1)

Notice that if y-1 = 3 (focal length) (or y=4), then

(x-1)^2 = 4*3*3 = 36

and x-1 = +/- 6, or x = 1 +/- 6 (latus rectum)

The focal length is 3, so the focus is 3 units above the vertex, at (1,4). When y=4, x = 1 +/- 6 (or x = -5 and x = 7), so the length of the latus rectum is 12. Notice that the latus rectum (12) is four times the focal length (3).

You now have 3 points: the vertex at (1,1) and the two ends of the latus rectum at (-5,4) and (7,4). That's enough to graph the parabola (although you might want to get a couple more points).

Answer 4

The vertex is at (1,1)
The axis of symmetry is the line x = 1
The distance between vertex and focus is the same as the distance from vertex to directrix. the equation of a parabola in terms of vertex, focus, and directrix is
(x - h)^2 = 4p(y - k)^2.
h = 1, k = 1, leaving 4p = 12, or p = 3, so the focus is at (1, 1 + 3), or (1,4), and the equation of the directrix is y = 1 - 3, or y = -2.
The direction of opening of the parabola is +y.
The latus rectum is the chord which passes through the focus parallel to the directrix, so solving the equation for y = 4,
(x - 1)^2 = 12(4 - 1) = 36
x = ± 6, so the length of the latus rectum is 6 + 6, or 12

Answer 5

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Answer 6

goto aol.homeworkhelp.com (maybe homework helper?)
do you have a friend you can copy off of?