The centre of the hyperbola $9 x^{2}-36 x-16 y^{2}+96 y-252=0$ is Ans $\times$ i. $(-2,-3)$
The centre of the hyperbola $9 x^{2}-36 x-16 y^{2}+96 y-252=0$ is Ans $\times$ i. $(-2,-3)$
- $(-2,-3)$
- $(2,-3)$
- $(-2,3)$
- $(2,3)$
Solution
$9 x^{2}-36 x-16 y^{2}+96 y-252=0$
$9(x-2)^{2}-16(y-3)^{2}=\frac{22}{532}$
center $\rightarrow(2,3)$
Asked in: MHT CET 2020 (20 Oct Shift 1)
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