rosedewittbukater
  • rosedewittbukater
Identify the type, center, and foci of the conic section. 11x^2 – 2y^2 + 66x – 16y + 45 = 0 (I'll post what I work I have done so far)
Mathematics
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chestercat
  • chestercat
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rosedewittbukater
  • rosedewittbukater
My work: \[11x ^{2}+66x-2y ^{2}-16y=-45\]\[11(x ^{2}+6x)-2(y ^{2}+8y)=-45\]\[11(x ^{2}+6x+(3^{2}))-2(y ^{2}+8y+(4^{2}))=-45+11(3^{2})+(-2)(4^{2})\]\[11(x ^{2}+6x+9)-2(y ^{2}+8y+16)=22\]\[11(x+3)^{2}-2(y+4)^{2}=22\]\[\frac{ 11(x+3)^{2} }{ 22 }-\frac{ (y+4)^{2} }{ 22 }=1\]\[\frac{ (x+3)^{2} }{ 11 }-\frac{ (y+4)^2 }{ 2 }=1\] I'm not sure how to tell what type of conic section it is. Can someone help me? Textbooks are not very helpful.
rosedewittbukater
  • rosedewittbukater
@whpalmer4 Do you know how to do this? If so, do you think you could help me?
rosedewittbukater
  • rosedewittbukater
@thomaster Are you good at these types of problems?

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rosedewittbukater
  • rosedewittbukater
@Mertsj Can you help me?
anonymous
  • anonymous
@rosedewittbukater when the sign is not the same you have a hyperbola. Because the \(x\)-term is positive, we have a *horizontal* hyperbola.
rosedewittbukater
  • rosedewittbukater
@oldrin.bataku how do you know its a hyparabola?
Mertsj
  • Mertsj
The negative sign between the two terms on the left tells you. If it were a plus sign, it would be an ellipse.
Mertsj
  • Mertsj
And the first denominator should be 2 not 11
rosedewittbukater
  • rosedewittbukater
Oh yeah, I wrote that wrong. That's what I meant to put in the denominator.
Mertsj
  • Mertsj
And the denominator of the y term should be 11
rosedewittbukater
  • rosedewittbukater
Umm not really that's confusing. @oldrin.bataku
rosedewittbukater
  • rosedewittbukater
So since it's a hyperbola, how do I find the center and foci?
Mertsj
  • Mertsj
\[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\]
Mertsj
  • Mertsj
That is the equation of a hyperbola whose center is (h,k). Make your equation look like that and figure out what (h,k) is. That is your center.
Mertsj
  • Mertsj
Then find c. Use: \[a^2+b^2=c^2\]
rosedewittbukater
  • rosedewittbukater
So the center is (-3, -4).
Mertsj
  • Mertsj
The foci are c units to the right and c units to the left of the center. And yes, your center is correct.
rosedewittbukater
  • rosedewittbukater
And the foci is (+/-12.85, 0)?
rosedewittbukater
  • rosedewittbukater
Wait I did the math wrong...
Mertsj
  • Mertsj
\[c^2=11+2=13\]
Mertsj
  • Mertsj
\[c=\sqrt{13}\]
Mertsj
  • Mertsj
Foci: \[(-3\pm \sqrt{13},-4)\]
rosedewittbukater
  • rosedewittbukater
Oh ok thank you so much!!! I really appreciate it. :)
Mertsj
  • Mertsj
yw

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