The latus-rectum of the hyperbola 16 x2-9 y2-144 isa 16 - 3b 32 - 3c 8 - 3d 4 - 3

(b) 32/3 The standard form of the hyperbola nbsp;16x2 9y2=144 is x29 y216=1. Here, nbsp;a2=9, #160;b2=16 Latus rectum nbsp;of the hyperbola = 2b2a=2 ...

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Standard XI

Mathematics

Sequence

The latus-rec...

Question

The latus-rectum of the hyperbola 16x² − 9y² = 144 is
(a) 16/3
(b) 32/3
(c) 8/3
(d) 4/3

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Solution

(b) 32/3

The standard form of the hyperbola $16 x^{2} - 9 y^{2} = 144$ is $\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$ .
Here, $a^{2} = 9, b^{2} = 16$
Latus rectum of the hyperbola = $\frac{2 b^{2}}{a} = \frac{2 \times 16}{3} = \frac{32}{3}$

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The latus-rectum of the hyperbola 16x2 − 9y2 = 144 is (a) 16/3 (b) 32/3 (c) 8/3 (d) 4/3

(b) 32/3 The standard form of the hyperbola 16x2-9y2=144 is x29-y216=1. Here, a2=9, b2=16 Latus rectum of the hyperbola = 2b2a=2×163=323

The latus-rectum of the hyperbola 16x² − 9y² = 144 is
(a) 16/3
(b) 32/3
(c) 8/3
(d) 4/3

(b) 32/3

The standard form of the hyperbola $16 x^{2} - 9 y^{2} = 144$ is $\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$ .
Here, $a^{2} = 9, b^{2} = 16$
Latus rectum of the hyperbola = $\frac{2 b^{2}}{a} = \frac{2 \times 16}{3} = \frac{32}{3}$