Question

Find the coordinates of foci for the hyperbola 9x² - 16y²-72x +96y - 144 = 0​

Answer 1

$\frac{(x - 4)^{2} }{ (\frac{12}{3})^{2} } - \frac{(y - 3)^{2} }{ (\frac{12}{4})^{2} } = 1$

Step-by-step explanation:

$Given, \: \:9x² - 16y²-72x +96y - 144 = 0 \\ 9x² - 16y²-72x +96y = 144 \\ 9x² - 16y²-72x +96y + 144 - 144 = 144\\ (9 {x}^{2} - 72x + 144) - (16 {y}^{2} - 96y + 144) = 144 \\ \{(3x)^{2} - 2 \times 3x \times 12 + (12)^{2} \} - \{ (4y)^{2} - 2 \times 4y \times 12 + (12)^{2} \} = 144 \\ (3x - 12)^{2} - (4y - 12)^{2} = 144 \\ \{3(x - 4) \}^{2} - \{ 4(y - 3)\}^{2} = 144\\ 9(x - 4)^{2} - 16(y - 3)^{2} = 144 \\ \frac{9(x - 4)^{2} - 16(y - 3)^{2} }{144} = \frac{144}{144} \\ \frac{9(x - 4)^{2}}{144} - \frac{16(y - 3)^{2}}{144} = 1 \\ \frac{(x - 4)^{2} }{ \frac{144}{9}} - \frac{(y - 3)^{2} }{ \frac{144}{16} } = 1 \\ \frac{(x - 4)^{2} }{ (\frac{12}{3})^{2} } - \frac{(y - 3)^{2} }{ (\frac{12}{4})^{2} } = 1$