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What is the centre of hyperbola y^2-4x^2-8x-12y+36=0

Option: 1

(1,6)


Option: 2

(1,-6)


Option: 3

(-1,6)


Option: 4

(-1,-6)


Answers (1)

best_answer

The given equation can be written as

 y^2-12y+36-4\left ( x^2+2x+1 \right )= -4

4(x+1)^2-(y-6)^2=4

(x+1)^2-\frac{1}{4}(y-6)^2=1

This is a shifted hyperbola with Centre at (-1,6)

Posted by

Ritika Jonwal

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