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Question

The equation of the hyperbola 4x232xy24y+24=0 in its standard form is

A
(x4)216- \dfrac{(y+2)^2}{36}=1$
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B
(x4)29- \dfrac{(y-2)^2}{36}=1$
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C
(x4)29- \dfrac{(y+2)^2}{36}=1$
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D
(x4)29- \dfrac{(y+2)^2}{16}=1$
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Solution

The correct option is C (x4)29- \dfrac{(y+2)^2}{36}=1$
4x232xy24y+24=0
4(x28x)(y2+4y)+24=0
4(x4)2(y+2)236=0
(x4)29(y+2)236=1

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