parabolas

[Guest]

I am sorry for posting so much I am just freaking out at the finals packet we just got.......

so my last 2 quations...for today ....

1) find the equation in standard form that has the vertex (3,2) and focus ( 3,4)

2) find the equation of the parabola in standard form that has focus (-2,2) and directrix x=4


the first one I have (x-3)^2= 8y-16

and the second one im not sure how to do because i dont know if i am doing the first one right

thanks again everyone

 

[soroban]

Hello, buyinuamonkey!

1) Find the equation in standard form that has the vertex (3,2) and focus ( 3,4)

I have: \(\displaystyle \,(x\,-\,3)^2\:=\:8(y\,-\,2)\;\;\) . . . Correct!

2) Find the equation of the parabola in standard form that has focus (-2,2) and directrix x=4

      F       |   V           :
    - * - - - + - * - - - - - +
   (-2,2)     | (1,2)       (4,2)
              |
              |               :
    --+-------+---+-----------+--
     -2       |   1           :4
              |               :

The focus is at \(\displaystyle F(-2,2)\); the directrix is the vertical line \(\displaystyle x\,=\,4\).

The vertex is halfway-between at \(\displaystyle V(1,2)\;\) . . . and we see that \(\displaystyle p\,=\,-3\)
\(\displaystyle \;\;\)(The direction from the Vertex to the Focus is to the left.)

This is a "horizontal" parabola with standard form: \(\displaystyle \y\,-\,k)^2\:=\:4p(x\,-\,h)\)

The equation is: \(\displaystyle \y\,-\,2)^2\;=\;-12(x\,-\,1)\)

 

[stapel]

Multi-post.