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Question 409759: Hi can you please help me with this question, graph the equation equal to:9x^2+16y^2+36x-32y-92=0 and find the vertix, focus point and axis of symmetry. Thank you all your help
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! 9x^2+16y^2+36x-32y-92=0
find the vertex, focus point and axis of symmetry
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9x^2+16y^2+36x-32y-92=0
factor and completing the square (to complete the square, coefficients of x^2 and y^2 must be=1
9(x^2+4x+4)+16(y^2-2y+1)=92+36+16=144
9(x+2)^2+16(y-1)^2=144
divide by 144
(x+2)^2/16+(y-1)^2/9=1
This is an ellipse with horizontal major axis and center at (-2,1)
standard form of an ellipse: (x-h)^2/a^2+(y-k)^2/b^2=1
a^2=16
a=4
b^2=9
b=3
c^2=a^2-b^2=16-9=7
c=sqrt(7)=2.65
The vertices are on the horizontal major axis, at x=-2+-a=-2+-4=-6 and 2
The focal points are also on the major axis, at x=-2+-c=-2+-sqrt(7)=-4.65 and .65. Axis of symmetry does not apply to ellipses
ans:
center:(-2,1)
vertices:(-6,1),(2,1)
foci:(-4.65,1),(.65,1)
axis of symmetry does not apply to ellipse. It applies to parabolas.
see the graph of the ellipse below:
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y=1+((144-9(x+2)^2)/16)^.5
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