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Ellipse question

2019 · 8 Apr · Shift 2 · Q26
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Ellipse question

2019 · 8 Apr · Shift 2 · Q26

JEE MainMathematicsEllipseMCQ+4 / −1
In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at (0,5 3\sqrt 33​), then the length of its latus rectum is :
  1. A
    5
  2. B
    8
  3. C
    10
  4. D
    6
View written solutionFree

Correct answer: A

  1. Set up the ellipse parameters

Since the centre is at the origin and one focus is at (0,53)(0,5\sqrt{3})(0,53​), the major axis is along the yyy-axis.

So the ellipse can be taken as x2b2+y2a2=1,a>b\frac{x^2}{b^2}+\frac{y^2}{a^2}=1, \qquad a>bb2x2​+a2y2​=1,a>b with foci at (0,±c),c2=a2−b2.(0,\pm c), \qquad c^2=a^2-b^2.(0,±c),c2=a2−b2.

Given one focus is at (0,53)(0,5\sqrt{3})(0,53​), we get c=53c=5\sqrt{3}c=53​ so c2=75.c^2=75.c2=75.

  1. Use the condition on axes lengths

Length of major axis =2a=2a=2a and length of minor axis =2b=2b=2b.

Their difference is given as 101010: 2a−2b=102a-2b=102a−2b=10 a−b=5.a-b=5.a−b=5.

  1. Use the focal relation

We know a2−b2=c2=75.a^2-b^2=c^2=75.a2−b2=c2=75. But a2−b2=(a−b)(a+b).a^2-b^2=(a-b)(a+b).a2−b2=(a−b)(a+b). So 5(a+b)=755(a+b)=755(a+b)=75 a+b=15.a+b=15.a+b=15.

Now solve a−b=5,a+b=15.a-b=5, \qquad a+b=15.a−b=5,a+b=15. Adding, 2a=20⇒a=10.2a=20 \Rightarrow a=10.2a=20⇒a=10. Then b=5.b=5.b=5.

  1. Find the length of the latus rectum

For an ellipse, length of latus rectum is 2b2a.\frac{2b^2}{a}.a2b2​. Thus \text{latus rectum} = \frac{2\cdot 5^2}{10}= rac{50}{10}=5.

  1. Compare with options

Thus the required length is 5.5.5. So the correct option is A.

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