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Let a parabola P have the equation y^2 = 4 5 (x + 5 ) . An ellipse E: x^2 a^2 + y^2 b^2 = 1 ( a > b ) is drawn such that the vertex of P coincides with one of the foci of E , and the latus rectum of P is the minor axis of E . The distance between the directrices of the ellipse E is equal to

Options

  1. A34 5
  2. B10 5
  3. C6 5
  4. D2 5

Correct answer

B. 10 5

Step-by-step solution

The equation of the parabola is y^2 = 4 5 (x + 5 ) . Comparing this with Y^2 = 4AX , we have A = 5 . The vertex of P is at (- 5 , 0) . The focus of P is at (- 5 + 5 , 0) = (0, 0) . The latus rectum of P is the line segment passing through its focus (0, 0) and perpendicular to its axis. Putting x = 0 in the parabola's equation gives y^2 = 4 5 ( 5 ) = 20 y = 2 5 . Thus, the latus rectum of P is the segment joining (0, 2 5 ) and (0, -2 5 ) . For the ellipse E: x^2 a^2 + y^2 b^2 = 1 , the foci are ( ae, 0) and the mino

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