anonymous
  • anonymous
Can someone help me with my conic sections assignment? I'm really confused.
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
I have a few that I really don't know how to do, but here's one: What is the equation of the following graph?
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phi
  • phi
First, what shape is that? is it a circle, ellipse, parabola or hyperbola?
anonymous
  • anonymous
ellipse

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phi
  • phi
next, look in your notes for the equation of an ellipse. can you write down the "general form" ?
anonymous
  • anonymous
\[\frac{ x^{2} }{ a ^{2}} + \frac{ y ^{2}}{ b ^{2} }\]
phi
  • phi
an equation has an = sign in it. it should be \[ \frac{ x^{2} }{ a ^{2}} + \frac{ y ^{2}}{ b ^{2} }=1 \] but that is for an ellipse whose center is at (0,0) your ellipse is not centered at (0,0). The better equation is \[ \frac{ (x-h)^{2} }{ a ^{2}} + \frac{ (y-k) ^{2}}{ b ^{2} } =1\] where (h,k) is the center of the ellipse. Can you figure out what point is in the exact center of your ellipse ?
anonymous
  • anonymous
(0,-2)
phi
  • phi
that means you know (h,k) h=0 and k= -2 put those into the equation
anonymous
  • anonymous
\[\frac{(x-0)^{2} }{ a ^{2} }+ \frac{ (y+2)^{2} }{ b ^{2}}\]
phi
  • phi
= 1 don't leave that out also, x-0 simplifies to x so \[ \frac{x^{2} }{ a ^{2} }+ \frac{ (y+2)^{2} }{ b ^{2}} =1 \]
phi
  • phi
now for the "a" value: that is the distance in the x-direction (sideways) from the center to the side. what is the value of "a" ?
anonymous
  • anonymous
1?
phi
  • phi
yes, a is 1 and a^2 which means a*a is 1*1 = 1 so \[ \frac{x^{2} }{ a ^{2} }+ \frac{ (y+2)^{2} }{ b ^{2}} =1 \\ \frac{x^{2} }{ 1 ^{2} }+ \frac{ (y+2)^{2} }{ b ^{2}} =1 \] now find b, which the distance from the center to the top (i.e. in the up/down direction , or y-axis direction)
anonymous
  • anonymous
3?
phi
  • phi
yes
phi
  • phi
\[ \frac{x^{2} }{ 1 ^{2} }+ \frac{ (y+2)^{2} }{ 3^{2}} =1 \] we could simplify that a little, and write \[ x^2 + \frac{ (y+2)^{2} }{ 9} =1 \] but either way is ok
anonymous
  • anonymous
So is that the final answer?
phi
  • phi
yes. You can check some points to make sure it's correct. for example , when x is 0 you get \[ 0^2 + \frac{ (y+2)^{2} }{ 9} =1 \\ \frac{ (y+2)^{2} }{ 9} =1 \\ (y+2)^2= 9 \\ y+2= \pm 3 \] and y= +3-2= 1 or y= -3-2= -5 so points (0,1) and (0, -5) should be on the ellipse.

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