The latusrectum of the hyperbola $9 x^{2} - 16 y^{2} + 72 x - 32 y - 16 = 0$ is _______________________.

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Solution

Given equation of hyperbola,

$9 x^{2} - 16 y^{2} + 72 x - 32 y - 16 = 0 i . e . 9 (x^{2} + 8 x) - 16 (y^{2} - 2 y) = 16 i . e . 9 (x^{2} + 2 (4) x + 16 - 16) - 16 (y^{2} - 2 y + 1 - 1) = 16 i . e . 9 [{(x + 4)}^{2} - 16] - 16 [{(y - 1)}^{2} - 1] = 16 i . e . 9 {(x + 4)}^{2} - 16 {(y - 1)}^{2} - 16 \times 9 + 16 = 16 i . e . 9 {(x + 4)}^{2} - 16 {(y - 1)}^{2} = 16 \times 9 i . e . \frac{{(x + 4)}^{2}}{16} - \frac{{(y - 1)}^{2}}{9} = 1 i . e . a^{2} = 16, b^{2} = 9 ∴ e^{2} = 1 + \frac{b^{2}}{a^{2}} i . e . e^{2} = 1 + \frac{9}{16} = \frac{25}{16} i . e . e = \frac{5}{4} then, latus rectum is \frac{2 b^{2}}{a} i . e . \frac{2 \times 9}{4} = \frac{9}{2}$

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