SOLUTION: Please help me solve this equation Find the Parabola properties and graph y^2+6x+6y=39 Vertex: Focus: Directrix: Axis of symmetry: Endpoints of latus retum:

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Please help me solve this equation Find the Parabola properties and graph y^2+6x+6y=39 Vertex: Focus: Directrix: Axis of symmetry: Endpoints of latus retum:       Log On


   

Question 1044782: Please help me solve this equation
Find the Parabola properties and graph
y^2+6x+6y=39

Vertex:
Focus:
Directrix:
Axis of symmetry:
Endpoints of latus retum:
Found 2 solutions by MathLover1, solver91311:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Find the Parabola properties and graph
given:
y%5E2%2B6x%2B6y=39
Vertex:
first write your equation in vertex form x-h=a%28y-k%29%5E2 (since y squared) where h and k are x and y coordinates of the vertex
6x=-y%5E2-6y%2B39
x=-y%5E2%2F6-6y%2F6%2B39%2F6
x=-%281%2F6%29%28y%5E2%2B6y%29%2B13%2F2
x=-%281%2F6%29%28y%5E2%2B6y%2Bb%5E2%29-%28-1%2F6%29b%5E2%2B13%2F2
x=-%281%2F6%29%28y%5E2%2B6y%2B3%5E2%29-%28-1%2F6%293%5E2%2B13%2F2
x=-%281%2F6%29%28y%2B3%29%5E2-%28-1%2F6%299%2B13%2F2
x=-%281%2F6%29%28y%2B3%29%5E2%2B3%2F2%2B13%2F2
x=-%281%2F6%29%28y%2B3%29%5E2%2B16%2F2
x=-%281%2F6%29%28y%2B3%29%5E2%2B8
k=-3 and h=8
vertex is at ( 8, -3)
Since the value of a is negative, the parabola opens left.
Focus:
The focus of a parabola can be found by adding p to the x-coordinate h if the parabola opens left or right.
(h%2Bp,k)
Find p, the distance from the vertex to the focus.
Find p, the distance from the vertex to a focus of the parabola by using the following formula.
p=1%2F4a
since a=-1%2F6, we have
p=1%2F%284%28-1%2F6%29%29
p=-6%2F4
p=-3%2F2
(8-3%2F2,-3)
(16%2F2-3%2F2,-3)
( 13%2F2, -3)->focus
Directrix:
The directrix of a parabola is the vertical line found by subtracting p from the x-coordinate h
of the vertex if the parabola opens left or right.
x=h-p
x=8-%28-3%2F2%29
x=16%2F2%2B3%2F2
x+=+19%2F2
Axis of symmetry:
Find the axis of symmetry by finding the line that passes through the vertex and the focus.
vertex is at ( 8, -3)
( 13%2F2, -3)->focus
and that will be y=-3
Endpoints of latus retum:
length of latus rectum = 4p=4%2A%28-3%2F2%29=2%28-3%29=-6
endpoints of the latus rectum: (h%2Bp,k-2p) and (h%2Bp,k%2B2p)
since p=-3%2F2, k=-3 and h=8 we have
(8%2B%28-3%2F2%29,-3-2%28-3%2F2%29) and (8%2B%28-3%2F2%29,-3%2B2%28-3%2F2%29)
(16%2F2-3%2F2,-3%2B3) and (16%2F2-3%2F2,-3-3)
(13%2F2,0) and (13%2F2,-6)




Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Since is the squared variable, your parabola is "sideways" and the general equation is:



The vertex is at the point and the directed distances from the vertex to the focus and the directrix are and .

The axis of symmetry is the horizontal line through the vertex, so

The focus is and the directrix is the vertical line

The latus rectum is the segment with endpoints and .

Now the trick is to get into the form





Now complete the square in the LHS:



Factor both sides:



Solve for



So, by inspection you have:

, , and .

You can do the rest of the arithmetic.

John

My calculator said it, I believe it, that settles it