Latus rectum of hyperbola 9{x^2} - 16{y^2} - 18x - 32y - 151 = 0 is

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10-2. Parabola, Ellipse, Hyperbola

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The latus rectum of the hyperbola $9{x^2} - 16{y^2} - 18x - 32y - 151 = 0$ is

A

$\frac{9}{4}$

B

$9$

C

$\frac{3}{2}$

D

$\frac{9}{2}$

Solution

(d) $9{x^2} – 18x + 9 – 16{y^2} – 32y – 16 = 144$

$ \Rightarrow \frac{{{{(x – 1)}^2}}}{{16}} – \frac{{{{(y + 1)}^2}}}{9} = 1$

==> Latus rectum $ = \frac{{2{b^2}}}{a} = \frac{{2 \times 9}}{4} = \frac{9}{2}$.

Standard 11

Mathematics

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