The vertices of the hyperbola (9x^{2} - 16y^{2} - 36x + 96y... | MATH - HYPERBOLA | SolveeIt

The vertices of the hyperbola

$9x^{2} - 16y^{2} - 36x + 96y - 252 = 0$ are

$(6,3)$ and $( - 6,3)$

$(6,3)$ and $( - 2,3)$

$( - 6,3)$ and $( - 6, - 3)$

None of these

Answer

$(6,3)$ and $( - 2,3)$

Explanation

Given Hyperbola is

$9x^{2} - 16y^{2} - 36x + 96y - 252 = 0$ ⇒

$9\left( x^{2} - 4x \right) - 16\left( y^{2} - 6y \right) - 252 = 0$ ⇒

$9(x - 2)^{2} - 36 - 16(y - 3)^{2} + 144 - 252 = 0$ ⇒

$9(x - 2)^{2} - 16(y - 3)^{2} = 144$

or $\frac{(x - 2)^{2}}{4^{2}} - \frac{(y - 3)^{2}}{3^{2}} = 1$

or Vertices $x - 2 = \pm 4$ & $y = - 3 = 0$

i.e. $2 \pm 4,3$ or $(6,3)\& ⥂ ( - 2,3$

Question: The vertices of the hyperbola \(9x^{2} - 16y^{2} - 36x + 96y - 252 = 0\) are...