SOLUTION: my problem reads, write the standard equation of each vertex, and these are the problems they give me
1. y^2-8y-8x+64=0
2. 36x^2-288x+25y^2-150y=99
Algebra.Com
Question 170131: my problem reads, write the standard equation of each vertex, and these are the problems they give me
1. y^2-8y-8x+64=0
2. 36x^2-288x+25y^2-150y=99
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
write the standard equation of each vertex, and these are the problems they give me
1. y^2-8y-8x+64=0
Complete the square on the y-terms:
y^2-8y+16 =8x-64+16
(y-4)^2 = 8x-48
(y-4)^2 = 8(x-6)
The vertex is (6,4)
---------------------------
2. 36x^2-288x+25y^2-150y=99
Complete the square on the x and on the y-terms:
36x^2-288x+? + 25y^2-150y+? = -99+?
36(x^2- 8x + 16) + 25(y^2 - 6y + 9) = -99 + 36*16 + 25*9
36(x-4)^2 + 25(y-3)^2 = 702
Divide thru by 702 to get:
(x-4)^2/19.5 + (y-3)^2/28.08 = 1
------------------------------
center at (4,3)
vertices at (4,3+sqrt(28.08)) and (4(3-sqrt(28.08))
=========================================
Cheers,
Stan H.
RELATED QUESTIONS
Ihave three problems that my father brought to my attention. H etried to do them and... (answered by Fombitz)
standard equation of the parabola... (answered by josgarithmetic)
im familar with writing conic sections in there standard equation but there are 2... (answered by stanbon)
Hi! I'm working on some Conic Section problems, and do ok when they are centered at zero. (answered by venugopalramana)
Please help me with these three problems.
Solve each equation by using the quadratic... (answered by jessicaherndon)
Write an equation of the parabola with its vertex at the origin if its focus is at (0,2)
(answered by lynnlo)
Write the standard equation for each circle. Find the radius and give the coordinates for (answered by scott8148)
Classify the conic section and write its equation in standard form.
1) 4x^2 + y^2 -... (answered by lynnlo)
Identify the conic, write its equation in standard form, and give the vertex/vertices and (answered by Edwin McCravy)