MATH! Urgent

umm suppose a parabola has a vertex at (0,2) and points (1,1)

how would I derive the equation and focus, i've been trying to understand this for so long, I can't get it. Does this parabola have the equation (y-2)^2 = x and a focus of 1/4?? Is that correct??

I searched Google under the key words "parabola equation focus" to get these possible sources:

http://en.wikipedia.org/wiki/Parabola
http://www.tpub.com/math2/13.htm
http://colalg.math.csusb.edu/~devel/precalcdemo/conics/src/parabola.html
http://www.algebralab.org/lessons/lesson.aspx?file=Algebra_conics_directrix.xml
http://mathworld.wolfram.com/Parabola.html

Use the <Find> command to locate keywords within these sites.

I hope this helps. Thanks for asking.

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

Vertex = (h, k)
Focus = (h, k + p)
Directrix: y = k - p

You are given the vertex (0,2) and a point (1,1).

Let's try to find p using the vertex and the point you were given:
(x - h)^2 = 4p(y - k)
(1 - 0)^2 = 4p(1 - 2)
1 = 4p(-1)
1 = -4p
-1/4 = p

Let's try to find the equation now that we know p:
(x - 0)^2 = 4(-1/4)(y - 2)
x^2 = -1(y - 2)
x^2 = -y + 2
Set the equation equal to 0:
x^2 + y - 2 = 0 -->this is the equation for the parabola.

(You can check the point you were given by substituting the values into the equation.)

Focus = (h, k + p)
Therefore: (0, 7/4) is the focus.

Directrix: y = k - p
Therefore: y = 9/4

In this case, the focus is below the directrix; therefore, the parabola opens downward and p is negative.

I hope this helps.

1. 👍 0
2. 👎 0
3. 👁 140

Similar Questions

1. Algebra 1 parabola's

I have a graph with a parabola on it and asked to write the function for f(x) in standard form. I know that standard is y=ax^2+bx+c obviously, since the points are on the graph we know x and y so a,b and c need to be found. The

asked by HELP on September 5, 2013
2. algebra

Suppose a parabola has vertex (-4,7) and also passes through the point (-3,8), write the equation of the parabola in vertex form.

asked by Lilah on November 2, 2015
3. algebra

Suppose a parabola has a vertex (-4,7) and also passes through the point (-3,8) Write the equation of the parabola in vertex form. f(x)=a(x-h)^2+k I believe h=-4 k=7 Not sure what to do from here.

asked by Lee on October 29, 2012
4. math help,algebra

Okay this is what i have to do but i think i am doing something wrong. directions are: Identify the axis of symmetry, create a suitable table of values, and sketch the graph (including the axis of symmetry). The problem is: y =

asked by jas20 on February 20, 2007
5. math

A parabola has the equation y = 4(x-3)^2-7 Choose 2 true statements: A) The parabola has a minimum value B) The parabola has a maximum value C) The parabola does not cross the y-axis D) The parabola does not cross the x-axis E)

asked by claudia on May 15, 2007
6. math

A parabola has the equation y = 4(x-3)^2-7 Choose 2 true statements: A) The parabola has a minimum value B) The parabola has a maximum value C) The parabola does not cross the y-axis D) The parabola does not cross the x-axis E)

asked by claudia on May 14, 2007
7. MATH

a. Write an equation compared to the equation of the standard parabola that satisfies the description of each parabola. 1. A parabola whose vertex is (0, -3) 2. A parabola whose vertex is (5, 1) 3. A parabola that opens down and

asked by NAMELESS on December 31, 2018
8. Math

y^2=8(x-3) how would you get the coordinates. vertex = 3,0 then what? let x=1, or -1, and solve for y. This is a parabola, if you are graphing, you will need several points to plot. vertex = (3,0) using x = 2 (2,4) (2,-4) > +3

asked by jim on November 2, 2006
9. MATH

a parabola passes via the points (1.5,1) and (3,-5). given that the vertex lies on the line 7x+3y-4=0,find the focus and the equation of the parabola

asked by KEV on May 18, 2012
10. pre cal

What is the center of the conic whose equation is x^2 + 2y^2 - 6x + 8y = 0 2.Which one of the following equations represents a hyperbola? (5 points) A) 3x^2 + y^2 + 12x - 7 = 0 B) 3x^2 + 3y^2 + 12x - 7 = 0 C) 3x^2 + y + 12x - 7 =

asked by Mecie on April 16, 2012

More Similar Questions