|
Question 246720: I have a math question with two parts. I am trying to work through the problem but am so confused. The question is as follows:
Part 1) P is the parabola with focus (3,1) and directrix x=7. I have placed on the graph the focus (3,1) and the directrix x = 7 and have come up with a vertex of (5,1). The equation I am using is x-h=a(y-k)^2. But I am confused as to what the true equation looks like.
Part 2) H is the hyperbola with a center at (1,1), an x-intercept at (2,0) and one vertex at (5/4,1). I am completely lost with this one.
Any help would be greatly appreciated.
Lori
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The question is as follows:
Part A) P is the parabola with focus (3,1) and directrix x=7.
Plot the focus and draw the directrix.
The vertex is half-way between the focus and the directrix,
so it is (5,1)
So h = 5, k = 1, p = -2
Form for a parabola opening to the left:
(y-k)^2 = 4p(x-h)
---
Your Problem:
(y-1)^2 = -8(x-5)
y^2-2y+1 = -8x+40
y^2-2y-39 = -8x
x = (-1/8)(y^2-2y-39)
----------------------------
Part B) H is the hyperbola with center at (1,1), an x-intercept at (2,0) and one vertex at (5/4,1).
Plot the two points:
h=1 , k=1
a = 5/4-1 = 1/4
Form:
(x-h)^2/a^2 - (y-k)^2/b^2 = 1
---
Your Problem:
(x-1)^2/(1/4)^2 - (y-1)^2/b^2 = 1
------------
Let (x,y) = (2,0)
(2-1)^2/(1/16) - (0-1)^2/b^2 = 1
16 - 1/b^2 = 1
1/b^2 = 15
b^2 = 1/15
---------------------------
Equation:
16(x-1)^2 - 15(y-1)^2 = 1
=============================
Cheers,
Stan H.
|
|
|
| |
