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Question 467310: Can someone please tell me which conic each of the following equations are? Thank you in advance!!!
a. 12x^2 = 29y^2 - y + 39
b. 12x^2 - 29y = x + 39
c. 29y^2 + 12x^2 = x +39
d. 12y^2 + 29y = 39 - 12x^2
I posted this question earlier, but someone answered with no answer so I'm reposting. Thank you.
Found 2 solutions by Alan3354, lwsshak3:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! a. 12x^2 = 29y^2 - y + 39
b. 12x^2 - 29y = x + 39
c. 29y^2 + 12x^2 = x +39
d. 12y^2 + 29y = 39 - 12x^2
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Get the x^2 and y^2 terms on the same side, then check them.
a. 12x^2 - 29y^2 = - y + 39
Hyperbola, the square terms are different signs.
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b. 12x^2 - 29y = x + 39
Parabola, only one x has a squared term
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c. 29y^2 + 12x^2 = x +39
Ellipse, the squared terms have the same sign, but different coefficients.
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d. 12y^2 + 29y - 39 + 12x^2 = 0
Circle, same sign, same coefficients.
A circle is actually a special case of ellipse, btw.
Answer by lwsshak3(11628)  (Show Source):
You can put this solution on YOUR website! Can someone please tell me which conic each of the following equations are? Thank you in advance!!!
a. 12x^2 = 29y^2 - y + 39
b. 12x^2 - 29y = x + 39
c. 29y^2 + 12x^2 = x +39
d. 12y^2 + 29y = 39 - 12x^2
...
To determine which conic the equations above represent, the best way is to complete the square first.
a. 12x^2 = 29y^2 - y + 39
12x^2-29y^2+ y=39
completing the square
12x^2-29(y^2-y/29+(1/58)^2)=39+29*(1/58)^2
12x^2-29(y-(1/58)^2=39+29*(1/58)^2
This is a hyperbola with a horizontal transverse axis with center at (0,1/58) (note the negative sign between the x and y terms and that the x-term comes first; if y-term comes first, then transverse axis is vertical)
Standard form for hyperbola: (x-h)^2/a^2-(y-k)^2/b^2=1
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b. 12x^2 - 29y = x + 39
29y=12x^2-x-39
divide by 29
y=(12/29)x^2-x/29-39/29
=12/29(x^2-x/12+(1/24)^2)-39/29-(1/24)^2
=12/29(x-1/24)^2)-(39/29+(1/24)^2)
This is a parabola which opens upwards with vertex at (1/24,-(39/29+(1/24)^2))
Standard form a parabola: y=A(x-h)^2+k
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c. 29y^2 + 12x^2 = x +39
29y^2+12x^2-x =39
29y^2+12(x^2-x/12+(1/24)^2) =39+12*(1/24)^2
29y^2+12(x-(1/24)^2=39+12*(1/24)^2
This is an ellipse with a vertical major axis and center at (1/24,0),(note how this differs from the equation of a hyperbola in that there is a plus sign between the x and y terms. If (x-h)^2 has the larger denominator, major axis is horizontal. Conversely, if (y-k)^2 has the larger denominator, major axis is vertical.
Standard form of ellipse: (x-h)^2/a^2+(y-k)^2/b^2=1, (a>b)
..
d. 12y^2 + 29y = 39 - 12x^2
12y^2+29y+12x^2=39
12(y^2+29y/12+(29/24)^2)+12x^2=39+12*(29/24)^2
12(y+29/24)^2+12x^2=39+12*(29/24)^2
divide by 12
(y+29/24)^2+x^2=[39+12*(29/24)^2]/12
This is a circle with center at (0,-29/24) and a radius=sqrt([39+12*(29/24)^2]/12)
standard form for circle (x-h)^2+(y-k)^2=r^2
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