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Hyperbola question

2022 · 25 Jul · Shift 2 · Q32
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  5. /2022 · 25 Jul · Shift 2 · Q32

Hyperbola question

2022 · 25 Jul · Shift 2 · Q32

JEE MainMathematicsHyperbolaMCQ+4 / −1
Let the foci of the ellipse x216+y27=1\frac{x^{2}}{16}+\frac{y^{2}}{7}=116x2​+7y2​=1 and the hyperbola x2144−y2α=125\frac{x^{2}}{144}-\frac{y^{2}}{\alpha}=\frac{1}{25}144x2​−αy2​=251​ coincide. Then the length of the latus rectum of the hyperbola is :
  1. A
    329\frac{32}{9}932​
  2. B
    185\frac{18}{5}518​
  3. C
    274\frac{27}{4}427​
  4. D
    2710\frac{27}{10}1027​
View written solutionFree

Correct answer: D

  1. Find the foci of the ellipse

The ellipse is

x216+y27=1\frac{x^2}{16}+\frac{y^2}{7}=116x2​+7y2​=1

So,

For an ellipse, the focal distance satisfies

c2=a2−b2=16−7=9c^2=a^2-b^2=16-7=9c2=a2−b2=16−7=9

Hence,

Therefore, the foci of the ellipse are

(±3,0)(\pm 3,0)(±3,0)
  1. Write the hyperbola in standard form

Given hyperbola:

x2144−y2α=125\frac{x^2}{144}-\frac{y^2}{\alpha}=\frac{1}{25}144x2​−αy2​=251​

Multiply both sides by 252525:

25x2144−25y2α=1\frac{25x^2}{144}-\frac{25y^2}{\alpha}=114425x2​−α25y2​=1

This becomes

x2144/25−y2α/25=1\frac{x^2}{144/25}-\frac{y^2}{\alpha/25}=1144/25x2​−α/25y2​=1

So for the hyperbola,

  1. Use the fact that foci coincide

For the hyperbola

x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1a2x2​−b2y2​=1

its foci are at

(±c,0),c2=a2+b2(\pm c,0),\quad c^2=a^2+b^2(±c,0),c2=a2+b2

Since the foci coincide with those of the ellipse,

Thus,

a2+b2=9a^2+b^2=9a2+b2=9

Substitute a2=14425a^2=\frac{144}{25}a2=25144​:

14425+b2=9\frac{144}{25}+b^2=925144​+b2=9 b2=9−14425=225−14425=8125b^2=9-\frac{144}{25}=\frac{225-144}{25}=\frac{81}{25}b2=9−25144​=25225−144​=2581​
  1. Length of latus rectum of the hyperbola

For the hyperbola

x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1a2x2​−b2y2​=1

the length of the latus rectum is

2b2a\frac{2b^2}{a}a2b2​

Here,

a=125,b2=8125a=\frac{12}{5},\quad b^2=\frac{81}{25}a=512​,b2=2581​

So,

Latus rectum=2⋅8125125=16225⋅512=2710\text{Latus rectum} = \frac{2\cdot \frac{81}{25}}{\frac{12}{5}} =\frac{162}{25}\cdot \frac{5}{12} =\frac{27}{10}Latus rectum=512​2⋅2581​​=25162​⋅125​=1027​
  1. Compare with options
2710\frac{27}{10}1027​

which is Option D.

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