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Question

A hyperbola having transverse axis of length 12 unit is confocal (having same foci) with the ellipse 3x2+4y2=12, then

A
equation of the hyperbola is x2y215=116
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B
eccenticity of the hyperbola is 4
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C
distance between the directrices of the hyperbola is 18 unit
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D
length of latus ractum of hyperbola is 152 unit
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Solution

The correct options are
A equation of the hyperbola is x2y215=116
B eccenticity of the hyperbola is 4
C distance between the directrices of the hyperbola is 18 unit
D length of latus ractum of hyperbola is 152 unit
Equation of ellipse is
3x2+4y2=12x24+y23=1
e=134=12
Foci are (±1,0)

Now hyperbola is also having same focus i.e. (±1,0).
If equation of hyperbola is
x2a2y2b2=1 and e1 is the eccentricity of hyberbola, then
2ae1=2
Given 2a=12
e1=4
and b2=a2(e211)=1516

So, the equation of the hyperbola is
x2116y21516=1x21y215=116

Distance between the directrices
=2ae1=18 unit

Length of latus ractum
=2b2a=152 unit

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