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Ellipse question

2025 · 4 Apr · Shift 1 · Q29
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Ellipse question

2025 · 4 Apr · Shift 1 · Q29

JEE MainMathematicsEllipseMCQ+4 / −1
The length of the latus-rectum of the ellipse, whose foci are (2,5)(2,5)(2,5) and (2,−3)(2,-3)(2,−3) and eccentricity is 45\frac{4}{5}54​, is
  1. A
    503\frac{50}{3}350​
  2. B
    185\frac{18}{5}518​
  3. C
    65\frac{6}{5}56​
  4. D
    103\frac{10}{3}310​
View written solutionFree

Correct answer: B

  1. Identify the center and focal distance

The foci are given as (2,5)(2,5)(2,5) and (2,−3)(2,-3)(2,−3).

Since both have the same xxx-coordinate, the major axis is vertical.

The center is the midpoint of the foci:

(2,5+(−3)2)=(2,1)\left(2,\frac{5+(-3)}{2}\right)=(2,1)(2,25+(−3)​)=(2,1)

The distance from the center to either focus is

c=∣5−1∣=4c=|5-1|=4c=∣5−1∣=4
  1. Use eccentricity relation

For an ellipse,

e=cae=\frac{c}{a}e=ac​

Given

e=45,c=4e=\frac{4}{5}, \quad c=4e=54​,c=4

so

45=4a\frac{4}{5}=\frac{4}{a}54​=a4​

Hence,

a=5a=5a=5
  1. Find b2b^2b2 using c2=a2−b2c^2=a^2-b^2c2=a2−b2

For an ellipse,

c2=a2−b2c^2=a^2-b^2c2=a2−b2

Thus,

b2=a2−c2=25−16=9b^2=a^2-c^2=25-16=9b2=a2−c2=25−16=9
  1. Length of latus rectum

The length of the latus rectum of an ellipse is

2b2a\frac{2b^2}{a}a2b2​

Substitute b2=9b^2=9b2=9 and a=5a=5a=5:

Length of latus rectum=2⋅95=185\text{Length of latus rectum}=\frac{2\cdot 9}{5}=\frac{18}{5}Length of latus rectum=52⋅9​=518​
  1. Match with options
185\frac{18}{5}518​

corresponds to Option B.

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