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Write an equation for an ellipse centered at the origin, which has foci at 
(0,+-24) and co-vertices at 
(+-10,0).

Write an equation for an ellipse centered at the origin, which has foci at (0,±24) (0, \pm 24) and co-vertices at (±10,0) ( \pm 10,0) .

Full solution

Q. Write an equation for an ellipse centered at the origin, which has foci at (0,±24) (0, \pm 24) and co-vertices at (±10,0) ( \pm 10,0) .
  1. Ellipse Orientation: We have:\newlineFoci: (0,±24)(0, \pm 24)\newlineCo-vertices: (±10,0)(\pm 10, 0)\newlineChoose the orientation of the ellipse.\newlineSince the foci are on the yy-axis, the ellipse is vertical.
  2. Identifying cc: We have:\newlineCenter (h,k)(h, k): (0,0)(0, 0)\newlineFoci: (0,±24)(0, \pm 24)\newlineIdentify the value of cc.\newlinec=(00)2+(240)2c = \sqrt{(0 - 0)^2 + (24 - 0)^2}\newline=02+242= \sqrt{0^2 + 24^2}\newline=576= \sqrt{576}\newline=24= 24
  3. Identifying bb: We have:\newlineCenter (h,k)(h, k): (0,0)(0, 0)\newlineCo-vertices: (±10,0)(\pm 10, 0)\newlineIdentify the value of bb.\newlineb=(100)2+(00)2b = \sqrt{(10 - 0)^2 + (0 - 0)^2}\newline=102+02= \sqrt{10^2 + 0^2}\newline=100= \sqrt{100}\newline=10= 10
  4. Calculating a: We know the relationship between aa, bb, and cc for an ellipse is c2=a2b2c^2 = a^2 - b^2. We have c=24c = 24 and b=10b = 10. Calculate the value of aa. a2=c2+b2=242+102=576+100=676a^2 = c^2 + b^2 = 24^2 + 10^2 = 576 + 100 = 676 a=a2=676=26a = \sqrt{a^2} = \sqrt{676} = 26
  5. Standard Form of the Ellipse: We know:\newline(h,k)=(0,0)(h, k) = (0, 0)\newlinea=26a = 26\newlineb=10b = 10\newlineWhat would be the standard form of the ellipse?\newline(x0)2/b2+(y0)2/a2=1(x - 0)^2/b^2 + (y - 0)^2/a^2 = 1\newlinex2/102+y2/262=1x^2/10^2 + y^2/26^2 = 1\newlinex2/100+y2/676=1x^2/100 + y^2/676 = 1

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