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MHT CET202619 April 2026Evening ShiftMathematicsEllipseActual

The eccentricity of the ellipse represented by the equation 7x^2 + 16y^2 - 14x + 64y - 377 = 0 is...

Options

  1. A3 4
  2. B7 4
  3. C1 2
  4. D3 8

Correct answer

A. 3 4

Step-by-step solution

7x^2 + 16y^2 - 14x + 64y - 377 = 0 7(x^2 - 2x) + 16(y^2 + 4y) = 377 7(x^2 - 2x + 1) + 16(y^2 + 4y + 4) = 377 + 7(1) + 16(4) 7(x - 1)^2 + 16(y + 2)^2 = 448 Dividing by 448 : (x - 1)^2 64 + (y + 2)^2 28 = 1 Comparing with the standard equation of an ellipse X^2 a^2 + Y^2 b^2 = 1 , we get a^2 = 64 and b^2 = 28 . The eccentricity e is given by e = 1 - b^2 a^2 e = 1 - 28 64 = 36 64 = 6 8 = 3 4 Answer: 3 4

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