Hello, gemidrea!
I'll assume you know the basic formulas for the parabola and the ellipse.
\(\displaystyle \;\;\)(If this is not true, please let me know.)
1) Given a parabola with vertex \(\displaystyle (\)-\(\displaystyle 3,5}\) and directrix \(\displaystyle x\,=\,-6\)
Find the focus.
Code:
: V |
: * |
: (-3,5) |
: |
- : - - - - - - - + - - - - - -
: |
x=-6
We know that the parabola opens to the right.
The vertex is: \(\displaystyle \,(h,k)\,=\,(-3,\,5)\) and we see that \(\displaystyle p\,=\,3\)
The focus is 3 units to the
right of the vertex: \(\displaystyle \,F(0,\,5)\)
2) Find the equation at the parabola with vertex \(\displaystyle (2,2)\) and focus \(\displaystyle (2,5)\)
Code:
|
| * F(2,5)
|
|
| * V(2,2)
|
- + - - - - - - - -
|
This parabola opens upward: \(\displaystyle \,(x\,-\,h)^2\:=\:4p(y\,-\,k)\)
The vertex is: \(\displaystyle \,(h,k)\,=\,(2,2)\) and we see that \(\displaystyle p\,=\,3\)
The equation is: \(\displaystyle \
x\,-\,2)^2\:=\:12(y\,-\,2)\)
3) Find the vertex, axis of symmetry, focus, and directrix for: \(\displaystyle \,x\,+\,1\:=\:\frac{1}{8}(y\,+\,3)^2\)
We have: \(\displaystyle \,(y\,+\,3)^2\:=\:8(x\,+\,1)\;\) . . the parabola opens to the right: \(\displaystyle \subset\)
The vertex is: \(\displaystyle \,V(-1,-3)\) and \(\displaystyle p\,=\,2\)
Code:
: |
- : - - - - - - + - -
: V | F
: * | *
: (-1,-3) | (1,-3)
x=-3
The directrix is 2 units to the
left of the vertex: \(\displaystyle x\,=\,-3\)
The focus is 2 units to the
right of the vertex: \(\displaystyle \,F(1,-3)\)
The axis of symmetry is the horizontal line: \(\displaystyle \,y\,=\,-3\)
4) Find the equation of the ellipse with x-intercepts \(\displaystyle 1,\;\sqrt{3}\) and y-intercepts \(\displaystyle \pm5\)
Are you sure this is the x-intercepts are correctly stated?
As written, it makes for a
very ugly problem!
5) Find the center, foci, and lengths of the major and minor axis of: \(\displaystyle \,x^2\,+\,4y^2\:=\:36\)
Divide by 36: \(\displaystyle \L\,\frac{x^2}{36}\,+\,\frac{y^2}{9}\:=\:1\)
The center is at: \(\displaystyle C(0,0)\), the origin.
We see that: \(\displaystyle \,a\,=\,6,\;b\,=\,3\)
Length of major axis: \(\displaystyle \,12\)
Length of minor axis: \(\displaystyle \,6\)
The foci are found at: \(\displaystyle \,c^2\:=\:a^2\,-\,b^2\)
So we have: \(\displaystyle \,c^2\:=\:36\,-\,9\:=\:27\;\;\Rightarrow\;\;c\,=\,\pm3\sqrt{3}\)
Therefore the foci are at: \(\displaystyle \,(\pm3\sqrt{3},0)\)
Edit: I had made a serious blunder on locating the foci.
\(\displaystyle \;\;\;\;\)I've corrected it . . .
