Advertisements
Advertisements
Question
The latus-rectum of the hyperbola 16x2 − 9y2 = 144 is
Options
16/3
32/3
8/3
4/3
Solution
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 = \[\frac{2 b^2}{a} = \frac{2 \times 16}{3} = \frac{32}{3}\]
APPEARS IN
RELATED QUESTIONS
Find the equation of the hyperbola satisfying the given conditions:
Foci (±4, 0), the latus rectum is of length 12.
Find the equation of the hyperbola satisfying the given conditions:
Vertices (±7, 0), e = `4/3`
Find the axes, eccentricity, latus-rectum and the coordinates of the foci of the hyperbola 25x2 − 36y2 = 225.
Find the equation of the hyperbola satisfying the given condition :
foci (± \[3\sqrt{5}\] 0), the latus-rectum = 8
find the equation of the hyperbola satisfying the given condition:
foci (± \[3\sqrt{5}\] 0), the latus-rectum = 8
find the equation of the hyperbola satisfying the given condition:
foci (0, ± 12), latus-rectum = 36
Write the eccentricity of the hyperbola whose latus-rectum is half of its transverse axis.
Write the length of the latus-rectum of the hyperbola 16x2 − 9y2 = 144.
If the latus-rectum through one focus of a hyperbola subtends a right angle at the farther vertex, then write the eccentricity of the hyperbola.
