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Question 766718: What is the width top to bottom and right to left of these two ellipses equation?
#1 x^2+9y^2-2x+36y+28=0
#2 5x^2+3y^2-10x-12y+2=0
Answer by DSMLMD(16)   (Show Source):
You can put this solution on YOUR website! To know the width top to bottom and right to left of ellipse, we must change those equation to standard ellipse equation. From the standard ellipse equation that we have changed, we can identify the width from vertex, both the major line and minor line. If the biggest value (a-value) is in x-fraction, the ellipse have a two long curve on left and right. If the biggest value is in y-fraction, the ellipse have a tow long curve on top and bottom.
#1
x^2+ 9y^2 - 2x + 36y + 28 = 0
x^2 - 2x + 9y^2 + 36y + 28 = 0
(x^2 - 2x + 1) + 9(y^2 + 4y) + 28 = 0
(x - 1)^2 - 1 + 9(y^2 + 4y + 4) + 28 = 0
(x - 1)^2 + 9(y + 2)^2 - 36 + 27 = 0
(x - 1)^2 + 9(y + 2)^2 - 9 = 0
(x - 1)^2 + 9(y + 2)^2 = 9
(x - 1)^2/9 + (y + 2)^2 = 1
(x - 1)^2/9 + (y + 2)^2/1 = 1
a^2 = 9 -> a = 3
b^2 = 1 -> b = 1
The long curve are in left and right.
The width top to bottom is 2b = 2
The width left to right is 2a = 6
#2
5x^2 + 3y^2 - 10x - 12y + 2 = 0
5x^2 - 10x + 3y^2 - 12y + 2 = 0
5(x^2 - 2x) + 3(y^2 - 4y) + 2 = 0
5(x^2 - 2x + 1) + 3(y^2 - 4y + 4) + 2 = 0
5(x - 1)^2 - 5 + 3(y - 2)^2 - 12 + 2 = 0
5(x - 1)^2 + 3(y - 2)^2 - 15 = 0
5(x - 1)^2 + 3(y - 2)^2 = 15
(x - 1)^2/3 + (y - 2)^2/5 = 1
a^2 = 5 -> a = √5
b^2 = 3 -> b = √3
The long curve are in top and bottom
The width top to bottom is 2a = 2√5
The width left to right is 2b = 2√3
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