Advertisements
Advertisements
Question
Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola.
16x2 – 9y2 = 576
Solution
Equation of hyperbola: 16x2 – 9y2 = 576
On dividing by 576,
`x^2/36 - y^2/64 = 1`
The transverse axis is along the x-axis.
a2 = 36, b2 = 64
∴ c2 = a2 + b2 = 36 + 64 = 100
∴ a = 6, b = 8, c = 10
The coordinates of the vertices are (±a, 0) or (±6, 0)
The coordinates of the foci are (±c, 0) or (±10, 0)
Eccentricity e = `c/a = 10/6 = 5/3`
Length of the latus rectum = `(2b^2)/a`
= `(2xx 64)/6`
= `64/3`
APPEARS IN
RELATED QUESTIONS
Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola.
`x^2/16 - y^2/9 = 1`
Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola.
`y^2/9 - x^2/27 = 1`
Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola.
9y2 – 4x2 = 36
Find the centre, eccentricity, foci and directrice of the hyperbola.
x2 − y2 + 4x = 0
Find the centre, eccentricity, foci and directrice of the hyperbola .
x2 − 3y2 − 2x = 8.
Find the eccentricity of the hyperbola, the length of whose conjugate axis is \[\frac{3}{4}\] of the length of transverse axis.
If the distance between the foci of a hyperbola is 16 and its ecentricity is \[\sqrt{2}\],then obtain its equation.
Write the eccentricity of the hyperbola 9x2 − 16y2 = 144.
Write the coordinates of the foci of the hyperbola 9x2 − 16y2 = 144.
Write the equation of the hyperbola of eccentricity \[\sqrt{2}\], if it is known that the distance between its foci is 16.
If the foci of the ellipse \[\frac{x^2}{16} + \frac{y^2}{b^2} = 1\] and the hyperbola \[\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}\] coincide, write the value of b2.
If e1 and e2 are respectively the eccentricities of the ellipse \[\frac{x^2}{18} + \frac{y^2}{4} = 1\]
and the hyperbola \[\frac{x^2}{9} - \frac{y^2}{4} = 1\] then write the value of 2 e12 + e22.
If e1 and e2 are respectively the eccentricities of the ellipse \[\frac{x^2}{18} + \frac{y^2}{4} = 1\] and the hyperbola \[\frac{x^2}{9} - \frac{y^2}{4} = 1\] , then the relation between e1 and e2 is
The equation of the conic with focus at (1, −1) directrix along x − y + 1 = 0 and eccentricity \[\sqrt{2}\] is
The eccentricity of the conic 9x2 − 16y2 = 144 is
The eccentricity of the hyperbola whose latus-rectum is half of its transverse axis, is
The eccentricity of the hyperbola x2 − 4y2 = 1 is
The distance between the foci of a hyperbola is 16 and its eccentricity is \[\sqrt{2}\], then equation of the hyperbola is
If the eccentricity of the hyperbola x2 − y2 sec2α = 5 is \[\sqrt{3}\] times the eccentricity of the ellipse x2 sec2 α + y2 = 25, then α =
The eccentricity the hyperbola \[x = \frac{a}{2}\left( t + \frac{1}{t} \right), y = \frac{a}{2}\left( t - \frac{1}{t} \right)\] is
The locus of the point of intersection of the lines \[\sqrt{3}x - y - 4\sqrt{3}\lambda = 0 \text { and } \sqrt{3}\lambda + \lambda - 4\sqrt{3} = 0\] is a hyperbola of eccentricity
If the latus rectum of an ellipse is equal to half of minor axis, then find its eccentricity.
Given the ellipse with equation 9x2 + 25y2 = 225, find the eccentricity and foci.
