$The latus-rectum of the hyperbola$

$16 x^{2} - 9 y^{2} = 144 i s$

16/3

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32/3

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8/3

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4/3

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Solution

The correct option is B
32/3

$The standard form of the hyperbola$

$16 x^{2} - 9 y^{2} = 144 i s$

$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$

$H e r e, a^{2} = 9, b^{2} = 16$

$L a t u s r e c t u m o f t h e h y p e r b o l a$

$= \frac{2 b^{2}}{a} = \frac{2 \times 16}{3} = \frac{32}{3}$

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Similar questions

Q. The latus rectum of the hyperbola

16 x^{2} - 9 y^{2} = 144

is-

Q. Write the length of the latus-rectum of the hyperbola 16x² − 9y² = 144.

Q. Find the eccentricity and length of the latus rectum of the hyperbola

16 x^{2} - 9 y^{2} = 144.

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

Find the lengths of transverse axis and conjugate axis, eccentricity, the co-ordinates of focus, vertices, length of the latus-rectum and equations of the directrices of the following hyperbola $16 x^{2} - 9 y^{2} = 144.$

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The latus-rectum of the hyperbola 16x2−9y2=144 is

The correct option is 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

$The latus-rectum of the hyperbola$

$16 x^{2} - 9 y^{2} = 144 i s$

The correct option is B
32/3

$The standard form of the hyperbola$

$16 x^{2} - 9 y^{2} = 144 i s$

$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$

$H e r e, a^{2} = 9, b^{2} = 16$

$L a t u s r e c t u m o f t h e h y p e r b o l a$

$= \frac{2 b^{2}}{a} = \frac{2 \times 16}{3} = \frac{32}{3}$