Conic sections - ellipse, parabola, hyperbola

[littleone]

My problem is:
Find an equation of the parabola that has vertex (1,2) and focus (-1,2).

I dont know how to solve this problem, its under the conic sections of my book. If anybody could help me solve this or give me some examples or refer me to other web pages where I can find examples I would really appreciate it. Thank you. :?

 

[stapel]

I dont know how to solve this problem, its under the conic sections of my book.

The book has a section on conic sections, but no explanations or examples...? So you're asking for lessons, to replace what the book omitted....?

Thank you!

Eliz.

 

[littleone]

Yes, the book has an example for this type of problem, but it's not helpful. I think it skips steps because when I try to follow the example I get lost and dont understand how they got where they got. I would really appreciate it if you had any tips or if you could refer me to some examples. Thank you

 

[stapel]

I would really appreciate it if you had any tips or if you could refer me to some examples.

My only "tip" would be to memorize the formulas, and to plug whatever you're given into them, hoping that this leads somewhere helpful. (It usually does!)

As for lessons with examples, there appear to be many. Find a few that look good to you, read the explanations, and study the worked examples, making sure you understand the steps they're taking as they work things out.

Have fun!

Eliz.

 

[nycfunction]

My problem is:
Find an equation of the parabola that has vertex (1,2) and focus (-1,2).

I dont know how to solve this problem, its under the conic sections of my book. If anybody could help me solve this or give me some examples or refer me to other web pages where I can find examples I would really appreciate it. Thank you. :?

Hello. I will try to take you through this question as best I can considering it is an online lesson. We are dealing with a parabola, which is NOT centered at the origin, which is the point (0, 0). This parabola has a vertex of (h, k) and NOT (0, 0). In other words, if you graph this parabola, the CENTER will not be the point (0, 0) on the xy-plane. Got it?

You were given two points:

One of the vertices (called a vertex), which is (1, 2) and one of the foci (called a focus), which is (-1, 2). The vertex and the focus both lie on the horizontal line y = 2 (this is our axis of symmetry). Where did 2 come from? Do you see that each point given has a 2 in the y-coordinate place? This means that y = 2 is the axis of symmetry of this parabola. The axis of symmetry is the line that cuts the parabola into two equal parts (cuts it in half). The line y = 2 is also PARALLEL to the x-axis. Are you with me so far?

We are looking for an equation for this parabola given the one of the vertices and one focus.

We need to find the distance a. The letter a is part of the general equation of a parabola that is not centered at the origin. So, how do we find the distance a?

We use the distance formula between two points.

Let sqrt = square root for short. Got it?

Distance formula is:

Distance = sqrt{(y2 - y1)^2 + (x2 - x1)^2}

I will use the coordinates given in the vertex and focus, plug them into the distance formula and simplify.

a = sqrt{2 - 2)^2 + (- 1- 1)^2}

a = sqrt{4}

a = 2

What does a = 2 mean? It means that this is the distance from the vertex given to you (1, 2) to the focus given to you (-1, 2).

Clear so far?

Also, because the focus lies to the right of the vertex as given to you in the question, we know that this parabola opens to the right. When the parabola opens to the right, here is the general equation of such a parabola:

(y - k)^2 = 4a(x - h)

This parabola with an axis of symmetry parallel to the x-axis, opens to the right when a > 0.

Let me ask you: Is a BIGGER THAN 0?

Yes, we found a to be 2 above. See it? Isn't 2 GREATER than 0? Yes it is!

Here are some things to know about this parabola that opens to the right and its general equation.

The vertex = (h, k )

What is the vertex given to you?

It is the point (1, 2), right?

We also know that a = 2, right?

We also know that a represents the distance from the vertex to the focus.

We take all this information and plug it into the equation (y - k)^2 = 4a(x - h) and simplify where needed.

Are you ready?

Here is our general equation again:

(y - k)^2 = 4a(x - h)

Vertex = (h, k) = (1, 2)...This means h = 1 and k = 2. See it?

What is a? We know that a = 2. We now plug and chug.

(y - 2)^2 = 4(2)(x -1), which becomes (y - 2)^2 = 8(x - 1).

Done!

The equation you needed to find is (y - 2)^2 = 8(x - 1).

I hope this helps.

P.S. Go to the link http://www.chalkdust.com/ and purchase the precalculus series on vhs or dvd. A professor teaches CONICS in simple language with plenty of lessons on the board, which you can see from the comfort of your own home on tv.