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Write an equation for an ellipse centered at the origin, which has foci at 
(0,+-sqrt55) and vertices at 
(0,+-sqrt89).

Write an equation for an ellipse centered at the origin, which has foci at (0,±55) (0, \pm \sqrt{55}) and vertices at (0,±89) (0, \pm \sqrt{89}) .

Full solution

Q. Write an equation for an ellipse centered at the origin, which has foci at (0,±55) (0, \pm \sqrt{55}) and vertices at (0,±89) (0, \pm \sqrt{89}) .
  1. Ellipse Orientation: We have:\newlineFoci: (0,±55)(0, \pm\sqrt{55})\newlineVertices: (0,±89)(0, \pm\sqrt{89})\newlineSince the foci and vertices are on the y-axis, the ellipse is vertical.
  2. Finding the Value of c: Identify the value of c (the distance from the center to a focus) using the coordinates of the foci.\newlinec=55c = \sqrt{55}
  3. Finding the Value of aa: Identify the value of aa (the distance from the center to a vertex) using the coordinates of the vertices.\newlinea=89a = \sqrt{89}
  4. Finding the Value of b: Use the relationship c2=a2b2c^2 = a^2 - b^2 to find the value of b2b^2 (the square of the distance from the center to a co-vertex).\newlinec2=a2b2c^2 = a^2 - b^2\newline(55)2=(89)2b2(\sqrt{55})^2 = (\sqrt{89})^2 - b^2\newline55=89b255 = 89 - b^2\newlineb2=8955b^2 = 89 - 55\newlineb2=34b^2 = 34
  5. Standard Form of the Equation: Now we have:\newlinea=89a = \sqrt{89}\newlineb=34b = \sqrt{34}\newlinec=55c = \sqrt{55}\newlineThe standard form of the equation for a vertical ellipse centered at the origin is:\newline(xh)2/b2+(yk)2/a2=1(x - h)^2/b^2 + (y - k)^2/a^2 = 1\newlineSince the center (h,k)(h, k) is at the origin (0,0)(0, 0), the equation simplifies to:\newlinex2/b2+y2/a2=1x^2/b^2 + y^2/a^2 = 1
  6. Substituting Values into the Equation: Substitute the values of aa and bb into the equation.\newlinex2(34)2+y2(89)2=1\frac{x^2}{(\sqrt{34})^2} + \frac{y^2}{(\sqrt{89})^2} = 1\newlinex234+y289=1\frac{x^2}{34} + \frac{y^2}{89} = 1

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