The length of the latus-rectum of the ellipse, whose foci are (2 ,5 ) and (2 ,-3 ) and eccentricity is 45, is

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Correct answer: B. (b) To find the length of the latus rectum of the ellipse, we first recognize that the foci of the ellipse are given as F_1 : (2 ,5 ) and F_2 : (2 ,-3 ). This indicates that the major axis is aligned along the y-axis. Calculate the distance between the foci: F_1F_2=8 The formula involving the distance between the foci and the eccentricity is: F_1F_2=2 be Here, the eccentricity e is given as 45. Thus, we can solve for b : 8=2b 45 ⟹b= 82 45=5 Determine a^2 : Using the relationship between eccentricity, semi-minor axis b, and semi-major axis a : e^2=1- a^2b^2=1- a^225= 1625 Solving for a^2 : 1- a^225= 1625 ⟹ a^225= 925 ⟹a^2=9 Thus, a=3. Compute the length of the latus rectum: The formula for the length of the latus rectum L is: L= 2a^2b Substituting the known values: L= 2 (9 )5= 185 Therefore, the length of the latus rectum of the ellipse is 185.

$The length of the latus-rectum of the ellipse, whose foci are (2, 5) and (2, - 3) and eccentricity is \frac{4}{5}, is$

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