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Let P be a parabola with vertex 2 , 3 and directrix 2 x + y = 6 . Let an ellipse E : x 2 a 2 + y 2 b 2 = 1 , a > b of eccentricity 1 2 pass through the focus of the parabola P . Then the square of the length of the latus rectum of E , is

Options

  1. A385 8
  2. B347 8
  3. C512 25
  4. D656 25

Correct answer

D. 656 25

Step-by-step solution

Let Z be the foot of perpendicular from vertex to directrix of parabola, Now, finding Z we get, ⇒ x - 2 2 = y - 3 1 = - 4 + 3 - 6 4 + 1 ⇒ x - 2 2 = y - 3 1 = - 1 5 ⇒ x - 2 2 = - 1 5 , y - 3 1 = - 1 5 ⇒ x = 12 5 , y = 16 5 ⇒ Z ≡ 12 5 , 16 5 Eccentricity of ellipse is given as 1 2 . Now, finding b 2 using eccentricity formula we get, ⇒ b 2 = a 2 1 - e 2 = a 2 2 So, equation of ellipse will be, ⇒ 144 25 a 2 + 256 25 × a 2 2 = 1 as x , y ≡ 12 5 , 16 5 ⇒ 144 25 a 2 + 512 25 a 2 = 1 ⇒ 656 25 a 2 = 1 ⇒ a 2 = 656 25 ⇒ b 2

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