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Question

For the hyperbola (3x−4y−12)2100−(4x+3y−12)2225=1

A
coordinates of center is(8425,1225)
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B
coordinates of center is (1225,8425)
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C
coordinates of foci is (84±1001325,12751325)
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D
coordinates of foci is (84±151325,12201325)
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Solution

The correct option is D coordinates of foci is (84±151325,12201325)
Here equation of the conjugate axis is 3x4y12=0
Equation of the transverse axis is 4x+3y12=0
So Centre will be the intersetion point of lines =(8425,1225)
Also e=132
Foci are at a distance of ae=13 units from the centre.
So, using parametric form of line, coordinates of foci are (8425±13cosθ,1225±13sinθ)
where tanθ=43
Required foci coordinates are (84±151325,12201325)

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