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Write an equation for an ellipse centered at the origin, which has foci at 
(0,+-15) and co-vertices at 
(+-8,0).

Write an equation for an ellipse centered at the origin, which has foci at (0,±15) (0, \pm 15) and co-vertices at (±8,0) ( \pm 8,0) .

Full solution

Q. Write an equation for an ellipse centered at the origin, which has foci at (0,±15) (0, \pm 15) and co-vertices at (±8,0) ( \pm 8,0) .
  1. Ellipse Orientation: We have:\newlineFoci: (0,±15)(0, \pm 15)\newlineCo-vertices: (±8,0)(\pm 8, 0)\newlineSince the foci are on the yy-axis, this indicates a vertical orientation for the ellipse.
  2. Identify cc: We have:\newlineCenter (h,k)(h, k): (0,0)(0, 0)\newlineFoci: (0,±15)(0, \pm 15)\newlineIdentify the value of cc (the distance from the center to a focus).\newlinec=(00)2+(150)2c = \sqrt{(0 - 0)^2 + (15 - 0)^2}\newline=02+152= \sqrt{0^2 + 15^2}\newline=225= \sqrt{225}\newline=15= 15
  3. Identify bb: We have:\newlineCenter (h,k)(h, k): (0,0)(0, 0)\newlineCo-vertices: (±8,0)(\pm 8, 0)\newlineIdentify the value of bb (the distance from the center to a co-vertex).\newlineb=(80)2+(00)2b = \sqrt{(8 - 0)^2 + (0 - 0)^2}\newline=82+02= \sqrt{8^2 + 0^2}\newline=64= \sqrt{64}\newline=8= 8
  4. Calculate aa: We know the relationship between aa, bb, and cc for an ellipse with a vertical orientation is c2=a2b2c^2 = a^2 - b^2. We have already found c=15c = 15 and b=8b = 8. Now we solve for aa. a2=c2+b2=152+82=225+64=289a^2 = c^2 + b^2 = 15^2 + 8^2 = 225 + 64 = 289 a=289=17a = \sqrt{289} = 17
  5. Standard Form: We know:\newline(h,k)=(0,0)(h, k) = (0, 0)\newlinea=17a = 17\newlineb=8b = 8\newlineThe standard form of the ellipse with a vertical orientation is:\newline(xh)2b2+(yk)2a2=1\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1\newlineSubstitute the values of hh, kk, aa, and bb:\newline(x0)282+(y0)2172=1\frac{(x - 0)^2}{8^2} + \frac{(y - 0)^2}{17^2} = 1\newlinex264+y2289=1\frac{x^2}{64} + \frac{y^2}{289} = 1

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