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Two chords of the ellipse x^2 + 3y^2 = 12 each have a slope of 1 and a length of 2 2 . If M₁ and M₂ are the mid-points of these chords and O is the origin, then the value of (OM₁)^2 + (OM₂)^2 is equal to:

Options

  1. A40 3
  2. B20 3
  3. C10 3
  4. D32 3

Correct answer

A. 40 3

Step-by-step solution

Let the equation of a chord with slope 1 be y = x + c . Substitute y = x + c into the equation of the ellipse x^2 + 3y^2 = 12 : x^2 + 3(x + c)^2 = 12 4x^2 + 6cx + 3c^2 - 12 = 0 Let the roots of this quadratic equation be x₁ and x₂ . The difference between the roots is: |x₁ - x₂| = (6c)^2 - 4(4)(3c^2 - 12) 4 = 36c^2 - 48c^2 + 192 4 = 192 - 12c^2 4 = 12 - 3 4 c^2 The length of the chord is given by |x₁ - x₂| 1 + m^2 , where m = 1 . Length = 12 - 3 4 c^2 2 = 24 - 3 2 c^2 Given that the length of the chord is 2 2 = 8 ,

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