Asked by B
For the equation 9x^2+4y^2+54x-8y+49=0 determine the center, vertices and foci?
Answers
Answered by
Steve
rearrange stuff and complete the squares:
9x^2+54x + 4y^2-8y = -49
9(x^2+6x) + 4(y^2-2y) = -49
9(x^2+6x+9) + 4(y^2-2y+1) = -49 + 9*9 + 4*1
9(x+3)^2 + 4(y-1)^2 = 36
(x+3)^2/4 + (y-1)^2/9 = 1
Now answer the questions.
9x^2+54x + 4y^2-8y = -49
9(x^2+6x) + 4(y^2-2y) = -49
9(x^2+6x+9) + 4(y^2-2y+1) = -49 + 9*9 + 4*1
9(x+3)^2 + 4(y-1)^2 = 36
(x+3)^2/4 + (y-1)^2/9 = 1
Now answer the questions.
Answered by
Ash
how do you determine vertices?
Answered by
Steve
for this ellipse, the major axis is vertical.
a = 3
b = 2
The vertices are at (-3,1±3)
The co-vertices are at (-3±2,1)
Looks like it's time to review your ellipses.
a = 3
b = 2
The vertices are at (-3,1±3)
The co-vertices are at (-3±2,1)
Looks like it's time to review your ellipses.
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