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Question 616210: For the conic section described by the equation 9x^2+4y^2-18x+16y-11=0
What is the center?
What is the length of the semimajor axes?
What is the focal length?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! For the conic section described by the equation 9x^2+4y^2-18x+16y-11=0
What is the center?
What is the length of the semimajor axes?
What is the focal length?
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9x^2+4y^2-18x+16y-11=0
complete the square
9x^2-18x+4y^2+16y-11=0
9(x^2-2x+1)+4(y^2+4y+4)=11+9+16
9(x-1)^2+4(y+2)^2=36
(x-1)^2/4+(y+2)^2/9=1
This is an equation for an ellipse with vertical major axis.
Its standard form: ((x-h)^2/b^2+(y-k)^2/a^2=1, (a>b), (h,k)=(x,y) coordinates of the center
For given equation:
center: (1,-2)
b^2=4
b=2
length of semi major or minor axis=2b=4
..
a^2=9
c^2=a^2-b^2=9-4=5
c=√5
focal length=22c=2√5
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