An archway will be constructed over a walkway. a piece of wood will need to be curved to match a parabola. explain to maurice how to find the equation of the parabola given the focal point and the directrix.
the directrix is at 9 and the focus is at (7, 6)
so the focus is on the opening side of the parabola
so for a vertex of (h,k)
opening up and down
p is the distance from vertex to focus
it is also the distance from the vertex to the closes point on directix
2p=distance from focus to directix
p is positive if the focus is above the directix
y=9 and (7,6)
directix is above the vertex
p is negative
from 9 to 6 is 3
p is negative
and the vertex will be 1.5 above the focus and 1.5 below the directix
vertex is (7,7.5)
(x-7)²=4(-1.5)(y-7.5) is de equation
The Law of Cosines is always preferable when there's a choice. There will be two triangle angles (between 0 and 180 degrees) that share the same sine (supplementary angles) but the value of the cosine uniquely determines a triangle angle.
To find a missing side, we use the Law of Cosines when we know two sides and their included angle. We use the Law of Sines when we know another side and all the triangle angles. (We only need to know two of three to know all three, because they add to 180. There are only two degrees of freedom, to answer a different question I just did.
2.An archway will be constructed over a walkway. A piece of wood will need to be curved to match a parabola. Explain to Maurice how to find the equation of the parabola given the focal point and the directrix.
We'll use the standard parabola, oriented in the usual way. In that case the directrix is a line y=k and the focus is a point (p,q).
The points (x,y) on the parabola are equidistant from the line to the point. Since the distances are equal so are the squared distances.
The squared distance from (x,y) to the line y=k is
The squared distance from (x,y) to (p,q) is
These are equal in a parabola:
Gotta go; more later if I can.
3.There are two fruit trees located at (3,0) and (−3, 0) in the backyard plan. Maurice wants to use these two fruit trees as the focal points for an elliptical flowerbed. Johanna wants to use these two fruit trees as the focal points for some hyperbolic flowerbeds. Create the location of two vertices on the y-axis. Show your work creating the equations for both the horizontal ellipse and the horizontal hyperbola. Include the graph of both equations and the focal points on the same coordinate plane.
4.A pipe needs to run from a water main, tangent to a circular fish pond. On a coordinate plane, construct the circular fishpond, the point to represent the location of the water main connection, and all other pieces needed to construct the tangent pipe. Submit your graph. You may do this by hand, using a compass and straight edge, or by using a graphing software program.
5.Two pillars have been delivered for the support of a shade structure in the backyard. They are both ten feet tall and the cross-sections of each pillar have the same area. Explain how you know these pillars have the same volume without knowing whether the pillars are the same shape.
left right or up down opening ones
left right ones are in form (y-k)²=4p(x-h)
up down ones are (x-h)²=4p(y-k)
in all of them, the vertex is (h,k)
p is the distance from the vertex to the focus, also the shortest distance from the vertex to the directix making p half of the distance of the shortest path from focus to directix
if p is positive, then the parabola opens up or right
if p is negative then the parabola opens down or left
if the directix is y=something, then it is a up down parabola
if directix is x=something, then it is a left right parabola
directix is outside the parabola, kind of at the back
so lets say we had
focus=(2,3) and directix is x=-4
dirextix is x =-4 so left right
from x=-4 to x=2 (focus), that is distance of 6
wait, ok, so directix is on oposite side of opening
hmm, -4 is to left of the 2 so
the direxix is at back so the parabola opens to the right
p=3, positive 3
then we have the vertex is halfway between those
so 3 back from focus is from (2,3) to (-1,3)
so vertex is (-1,3)
draw a perpendicular line from the directrix passing through the focus, this will be the line of symmetry. The vertex(h, k) will be located on the line half way between the focus and directrix. The distance from the focus to the vertex is called the focal length, call it a. The then equation is (x - h)^2 = 4a(y - k) the equation can be manipulated to y = 1/4a(x - h)^2 + k hope it helps
To solve the equation of the parabola, I need to use the distance formula and substitute the focal point in (x1-y1). Also, I will use the distance formula for the direct x. Since it is a straight line, the x part could be eliminated and the remaining part of the formula would be used, substituting the y for the given measurement of the direct x. The two unfinished formulas would be combined and then solved for the equation. f(x) would be the unit of measurement for the parabola.
Hope this helps!!! :)
the vertex is (h,k)
given that the the vertex is (-2,-20)
when x=0, then y=-12
if the last term is positive, then maybe y=2(x+2)^2+(-20)?
unless you typed something wrong, then you are right that the last term is -20
you're missing vital information, add to it so we can solve this problem for you.
square root means have of a number so have of 3 is 1.5 then multiply it by 8 and you get your answer!