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Hyperbola question

2003 · Shift 0 · Q114
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Hyperbola question

2003 · Shift 0 · Q114

JEE MainMathematicsHyperbolaMCQ+4 / −1
The foci of the ellipse x216+y2b2=1{{{x^2}} \over {16}} + {{{y^2}} \over {{b^2}}} = 116x2​+b2y2​=1 and the hyperbola x2144−y281=125{{{x^2}} \over {144}} - {{{y^2}} \over {81}} = {1 \over {25}}144x2​−81y2​=251​ coincide. Then the value of b2{b^2}b2 is :
  1. A
    999
  2. B
    111
  3. C
    555
  4. D
    777
View written solutionFree

Correct answer: D

  1. Find the foci of the hyperbola

Given

x2144−y281=125\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}144x2​−81y2​=251​

First convert it to standard form by multiplying both sides by 252525:

25x2144−25y281=1\frac{25x^2}{144}-\frac{25y^2}{81}=114425x2​−8125y2​=1

So,

x2144/25−y281/25=1\frac{x^2}{144/25}-\frac{y^2}{81/25}=1144/25x2​−81/25y2​=1

Hence for the hyperbola,

a2=14425,b2=8125a^2=\frac{144}{25}, \qquad b^2=\frac{81}{25}a2=25144​,b2=2581​

For a hyperbola of form

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

its foci are at

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

Therefore,

c2=14425+8125=22525=9c^2=\frac{144}{25}+\frac{81}{25}=\frac{225}{25}=9c2=25144​+2581​=25225​=9

so

c=3c=3c=3

Thus the hyperbola has foci

(±3,0)(\pm 3,0)(±3,0)
  1. Use the same foci for the ellipse

The ellipse is

x216+y2b2=1\frac{x^2}{16}+\frac{y^2}{b^2}=116x2​+b2y2​=1

Since the foci are on the xxx-axis, we must have 16>b216>b^216>b2, so here

a2=16,minor axis parameter=b2a^2=16, \qquad \text{minor axis parameter}=b^2a2=16,minor axis parameter=b2

For an ellipse of form

x2a2+y2b2=1(a2>b2),\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad (a^2>b^2),a2x2​+b2y2​=1(a2>b2),

its foci are at

(±c,0),c2=a2−b2(\pm c,0), \qquad c^2=a^2-b^2(±c,0),c2=a2−b2

Given the foci coincide with those of the hyperbola, we have

c=3⇒c2=9c=3 \Rightarrow c^2=9c=3⇒c2=9

So,

9=16−b29=16-b^29=16−b2

which gives

b2=16−9=7b^2=16-9=7b2=16−9=7
  1. Check the options

The correct value is

b2=7b^2=7b2=7

So the correct option is D.

Previous

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