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Question

# The equation of hyperbola whose coordinates of the foci are (±8,0) and the length of latus rectum is 24 units, is

A
3x2y2=48
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B
4x2y2=48
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C
x23y2=48
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D
x24y2=48
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Solution

## The correct option is A 3x2−y2=48Let equation of hyperbola bex2a2−y2b2=1 ......(i)Given, foci, (±8,0)=(±ae,0)⇒ae=8 .......(ii)and length of latusrectum =2b2a∴24=2b2a⇒b2=12a ........(iii)∴ From equation (ii),a2e2=64⇒a2(a2+b2a2)=64⇒a2+b2=64⇒a2+12a=64⇒a2+12a−64=0⇒a2+16a−4a−64=0⇒a(a+16)−4(a+16)=0⇒(a+16)(a−4)=0⇒a=4 [∵ a cannot be negative]On putting a=4 in equation (iii), we getb2=12×4⇒b2=48∴ From equation (i),x216−y248=1⇒3x2−y2=48

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