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Maths-

General

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Question

The equation of a hyperbola, whose foci are (5, 0) and (–5, 0) and the length of whose conjugate axis is 8, is

$9 x^{2} - 16 y^{2} = 144$
$16 x^{2} - 9 y^{2} = 144$
$9 x^{2} + 16 y^{2} = 144$
$16 x^{2} - 9 y^{2} = 12$

The correct answer is: $16 x to the power of 2 end exponent minus 9 y to the power of 2 end exponent equals 144$

$b equals 4$ Þ $2 a e equals 10$ Þ $16 equals 25 minus a to the power of 2 end exponent$ Þ $a equals 3$
Hence the hyperbola is $16 x to the power of 2 end exponent minus 9 y to the power of 2 end exponent equals 144$ .

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Chemistry-General

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Chemistry-General

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