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Question 618147: Identify the conic section. If it is an ellipse or a hyperbola, give the center and foci.
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Found 2 solutions by solver91311, lwsshak3:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
The equation has both  and  terms, so it is not a parabola.
The equation has the same sign on the and terms, so it is not a hyperbola.
The equation has different valued coefficients on the and terms, so it is not a circle.
Hence, the equation describes an ellipse.
Complete the square on both the  and  terms, then divide both sides by the constant term remaining in the RHS so that the form looks like:
^2}{a^2}\ +\ \frac{(y\,-\,k)^2}{b^2}\ =\ 1)
Since the coefficient on is smaller than the coefficient on , the denominator below the ^2) will be the larger denominator and since there is no  term, the major axis of the ellipse will be parallel to the -axis.
So, center at
)
Vertices at:
)
where  is the square root of the denominator beneath .
Foci at:
)
where  , is the square root of the denominator beneath and  is the square root of the denominator beneath ^2)
John
My calculator said it, I believe it, that settles it
Answer by lwsshak3(11628)  (Show Source):
You can put this solution on YOUR website! Identify the conic section. If it is an ellipse or a hyperbola, give the center and foci.
4x^2+7y^2+32x-56y+148=0
complete the square:
4x^2+32x+7y^2-56y=-148
4(x^2+8x+16)+7(y^2-8y+16)=-148+64+112
4(x+4)^2+7(y-4)^2=28
divide by 28
(x+4)^2/7+(y-4)^2/4=1
This is an equation of an ellipse with horizontal major axis.
equation: (x-h)^2/a^2+(y-k)^2/b^2=1, a>b, (h,k)=(x,y) coordinates of center
For given ellipse:
center: (-4,4)
a^2=7
a=√7≈2.6
b^2=4
b=√4=2
c^2=a^2-b^2=7-4=3
c=√3≈1.7
Foci:(-4±c,4)=(-4±√3,4)=(-4±1.7,4)=(-5.7,4) and (-2.3,4)
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