Let the point L lying in the first quadrant be one end of a latus rectum of the ellipse x^2 4 + y^2 3 =1 . Let P and Q be the points where the normal drawn at L to this given ellipse meets the major axis and the minor axis. Then the distance between P and Q is

Step-by-step solution

Given ellipse is x^2 4 + y^2 3 =1 Here, a=2, b= 3 Now, e= 1- b^2 a^2 = 1- 3 4 = 1 2 L is lying in the first quadrant L = (a e, b^2 a )= (2 1 2 , 3 2 )= (1, 3 2 ) Now, y^ =- b^2 a^2 , x y =- 3 4 1 3 2 =- 1 2 m=- 1 y^ =2 Equation of normal at L is aligned & (y- 3 2 )=2(x-1) 2 x-y- 1 2 =0 & x 1 4 + y (- 1 2 ) =1 aligned Which meet to the minor axis at (0,- 1 2 ) and major axis at ( 1 4 , 0 ) . P ( 1 4 , 0 ) & Q (0,- 1 2 ) Distance between P & Q is D= ( 1 4 )^2+ ( 1 2 )^2 = 1 16 + 1 4 = 5 4