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Write an equation for an ellipse centered at the origin, which has foci at (+-5,0) and vertices at (+-sqrt41,0).

Write an equation for an ellipse centered at the origin, which has foci at (±5,0) ( \pm 5,0) and vertices at (±41,0) ( \pm \sqrt{41}, 0) .

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Q. Write an equation for an ellipse centered at the origin, which has foci at (±5,0) ( \pm 5,0) and vertices at (±41,0) ( \pm \sqrt{41}, 0) .
  1. Given Data: We are given the foci at (±5,0)(\pm 5, 0) and vertices at (±41,0)(\pm\sqrt{41}, 0). Since the foci and vertices are on the x-axis, this indicates a horizontal orientation for the ellipse.
  2. Calculate aa: The distance from the center to a vertex is represented by aa, and the distance from the center to a focus is represented by cc. Using the given vertices, we can find the value of aa.a=41a = \sqrt{41}
  3. Calculate 'c': Using the given foci, we can find the value of 'c'.\newlinec=5c = 5
  4. Use Relationship Equation: The relationship between aa, bb, and cc for an ellipse is given by the equation c2=a2−b2c^2 = a^2 - b^2. We can use this to find the value of bb.
    c2=a2−b2c^2 = a^2 - b^2
    52=(41)2−b25^2 = (\sqrt{41})^2 - b^2
    25=41−b225 = 41 - b^2
    b2=41−25b^2 = 41 - 25
    b2=16b^2 = 16
  5. Find 'b': Now we can find the value of 'b' by taking the square root of b2b^2.b=16b = \sqrt{16}b=4b = 4
  6. Write Standard Form: With the values of aa and bb, we can write the standard form equation of the ellipse. Since the ellipse is horizontally oriented and centered at the origin, the standard form is:\newline(x2a2)+(y2b2)=1(\frac{x^2}{a^2}) + (\frac{y^2}{b^2}) = 1
  7. Substitute Values: Substitute the values of aa and bb into the equation.(x2412)+(y242)=1(\frac{x^2}{\sqrt{41}^2}) + (\frac{y^2}{4^2}) = 1(x241)+(y216)=1(\frac{x^2}{41}) + (\frac{y^2}{16}) = 1

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