The length of the latus rectum of an ellipse is 6 units and the distance between a focus and its nearest vertex on the major axis is 53 units. If e is the eccentricity of this ellipse, then e satisfi…

Question

The length of the latus rectum of an ellipse is 6 units and the distance between a focus and its nearest vertex on the major axis is 53 units. If e is the eccentricity of this ellipse, then e satisfies the equation

Accepted Answer

Correct answer: A. (a) Given length of latns rectum is 6 units. Then , 2b^2a=b ⇒b^2=3a … . . (i ) vertex is (a ,0 ) and foci is (c, 0 ). According to question, a-c= 53 a- 53=ae Take square both side, a^2+ 259- 1099=a^2e^2 Here, e^2=1- b^2a^2 e^2=1- 3aa^2 from (i ) e^2=1- 3a Then, a^2+ 25a- 1093=a^2(1- 3a) 9a^2-30a+25=9a^2-27aa= 253 From (i), b^2=3 253=25b= 5c^2=a^2-b^2= 6259-25= 625-2259= 4009c= 203 Then, e= ca= ( 203 ) ( 253 )= 45 Now, check by option (A) with equation. 25x^2-40x+16=0 ⇒5x (5x-4 )-x (5x-4 )=0 ⇒ (5x-4 )^2=0x=4/5 Its root is equal to eccentricity, then this equation satisfy the eccentricity (e). So, option (a) is correct.

$The length of the latus rectum of an ellipse is 6 units and the distance between a focus and its nearest vertex on the major axis is \frac{5}{3} units. If e is the eccentricity of this ellipse, then e satisfies the equation$

Solution

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