The latus rectum of the hyperbola $16 x^{2} - 9 y^{2} = 144$ is-

\frac{13}{6}

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\frac{32}{3}

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\frac{8}{3}

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\frac{4}{3}

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Solution

The correct option is C $\frac{32}{3}$
We have,

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

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

Here, $a = 3, b = 4$

We know that the latus rectum

$= \frac{2 b^{2}}{a}$

Therefore,

$= \frac{2 \times 16}{3}$

$= \frac{32}{3}$

Hence, this is the answer.

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Similar questions

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

$The latus-rectum of the hyperbola$

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

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.

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 C 323We have,16x2−9y2=144x29−y216=1Here, a=3,b=4We know that the latus rectum =2b2aTherefore,=2×163=323Hence, this is the answer.

The latus rectum of the hyperbola $16 x^{2} - 9 y^{2} = 144$ is-

The correct option is C $\frac{32}{3}$
We have,

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

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

Here, $a = 3, b = 4$

We know that the latus rectum

$= \frac{2 b^{2}}{a}$

Therefore,

$= \frac{2 \times 16}{3}$

$= \frac{32}{3}$

Hence, this is the answer.