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Question

If the foci of the ellipsex216+y2b2=1and the hyperbola x2144y281=125coincide,write the value of b2

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Solution

The given equation of hyperbola is

x2144y281=125

25x214425y281=1

x2144125y28125=1

x2(125)2y2(95)2=1

This equation of the hyperbola is of the fromx2a21y2b21=1,

where a21=14425 and b21=8125

Let,e,be the eccentricity of the hyperbola.Then,

e1=1+b21a21

=1+812514425

=1+81144

=144+81144

=1512

=54

So,the coordinates offoci are (±a1e1,0)i.e.,(±3,0)

It is given that the foci of the elipse coincide with the foci of the hyperbola.So,the coordinates of foci of the hyperbola are(±3,0)

Now,the given equation of ellipse is

x216+y2b2=1

x242+y2b2=1

This equation of the hyperbola is of theformx2a2y2b2=1

Where a22=16 and b22=b2

Let,e2 be the eccentricity of the given ellipse.So,the coordinates of foci are (±a2e2.0)

a2e2=3

4×e2=3 (a2=4)

e2=34

e22=916

Now,

b22=a22(1e22)

=16(1916)

=16×716

b22=7 (b22b2)

b2=7

Hence,b2=7


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