conic test help please?

Question

I seem to be clueless on almost all of it if any one could possibly do a few from each section it would be of great help to me. Thank you

For each parabola, determine: a)how it opens b)width c)vertex d)axis e)focus f)directrix g)y-int or x-int.
1)2x+6y^2=0
a)left
2)-8y=-x^2
a)up
3)y=-1/3(x-2)^2+7
a)down
4)x=5(y+3)^2-4
a)right
5)y+1=-1/4(x-3)^2
a)down
6)y^2-4x-2y=3
a)right
Write the equation for the parabola with:
7)vertex(0,0); focus(-5,0)

8)vertex:(0,0);directrix; x=-2

9)focus(6,0); y-6

Find the a. center and b. radius of each circle
10) x^2+(y+3)^2=36
(0,0); 6
11)(x+5)^2+(y-2)^2=10
(-5,2); sqr. of 10
12)10x^2+10y^2=70
(0,0);sqr. 7
13)4x^2=4y^2=20x-16y+37=0

Write the equation of a circle with:
14)center:(-2,-1); radius= sqr. of 8
(x+2)^2+(y+1)^2=8
For each ellipse find a) center b)vertices c)co. vertices d)foci
15)49x^2+y ^2=49

16)9x^2+25y^2=225

17)(x+3)^2/25 + (y+2)^2/49=1

18)3x^2+6x+5y^2-20y-13=0

Write the equation of an ellipse with:
19)foci(+or -5, 0) co. vertices (0, +or-2)

20)height 20 width 28 center (0,0)

21)center (0.-2) vertical major axis length 6, minor axis length 2

Answer 1

One problem per question please.

This is representative of a lot of them.

5) y + 1 = -1/4(x - 3)²

The x term is squared so it is a vertical parabola.

The coefficient of the x squared term is negative so it opens down.

y + 1 = -1/4(x - 3)²

The vertex (h, k) = (3, -1).

Since the parabola is vertical, the line of symmetry is vertical and runs thru the vertex. The line of symmetry is:

x = 3

Now find the focus and directrix.

p = directed distance from vertex to focus

1/(4p) = -1/4
p = -1

The focus is also on the line of symmetry and is:

(h, k + p) = [3, -1+(-1)] = (3, -2)

The directrix is a line perpendicular to the line of symmetry and is therefore horizontal. It is a directed distance -p from the vertex. Its equation is:

y = k - p = -1 - (-1) = 0
y = 0

Find the intercepts.

To find the x-intercepts let y = 0 and solve for x.

0 + 1 = -1/4(x - 3)²
-4 = (x - 3)²

This is not possible since a square cannot be negative. There are no x-intercepts.

To find the y-intercept let x = 0 and solve for y.

y + 1 = -1/4(0 - 3)² = -9/4
y = - 9/4 - 1 = - 13/4

The y-intercept is (0, -13/4).

I don't know what you mean by "width". That is an ambiguous term for a parabola. Perhaps you are referring to the latus rectum. It is the line segment perpendicular to the line of symmetry that starts on one side of the parabola, goes thru the focus and extends to the other side of the parbola. Its length is |4p| = 4. It extends from:

(h + 2p, k) to (h - 2p, k)
(3 + 2(-1), -1) to (3 - 2(-1), -1)
(1, -1) to (5, -1)