Advertisements
Advertisements
प्रश्न
The equation of the hyperbola with vertices at (0, ± 6) and eccentricity `5/3` is ______ and its foci are ______.
उत्तर
The equation of the hyperbola with vertices at (0, ± 6) and eccentricity `5/3` is `y^2/36 - x^2/64` and its foci are (0, ± 10).
Explanation:
Let equation of the hyperbola is `- x^2/a^2 + y^2/b^2` = 1
Vertices are (0, ± b)
∴ b = 6 and e = `5/3`
We know that e = `sqrt(1 + a^2/b^2)`
⇒ `5/3 = sqrt(1 + a^2/36)`
⇒ `25/9 = 1 + a^2/36`
⇒ `a^2/36 = 25/9 - 1 = 16/9`
⇒ `a^2 = 16/9 xx 36`
⇒ `a^2` = 64
So the equation of the hyperbola is `(-x^2)/64 + y^2/36` = 1
⇒ `y^2/36 - x^2/64` = 1
And foci = (0, ± be) = `(0, +- 6 xx 5/3)` = (0, ± 10)
APPEARS IN
संबंधित प्रश्न
Find the equation of the hyperbola satisfying the given conditions:
Vertices (0, ±5), foci (0, ±8)
Find the equation of the hyperbola satisfying the given conditions:
Vertices (0, ±3), foci (0, ±5)
Find the equation of the hyperbola satisfying the given conditions:
Foci (±5, 0), the transverse axis is of length 8.
Find the equation of the hyperbola satisfying the given conditions:
Foci (0, ±13), the conjugate axis is of length 24.
Find the equation of the hyperbola whose focus is (0, 3), directrix is x + y − 1 = 0 and eccentricity = 2 .
Find the equation of the hyperbola whose focus is (1, 1) directrix is 2x + y = 1 and eccentricity = \[\sqrt{3}\].
Find the eccentricity, coordinates of the foci, equation of directrice and length of the latus-rectum of the hyperbola .
9x2 − 16y2 = 144
Find the eccentricity, coordinates of the foci, equation of directrice and length of the latus-rectum of the hyperbola .
16x2 − 9y2 = −144
Find the eccentricity, coordinates of the foci, equation of directrice and length of the latus-rectum of the hyperbola .
3x2 − y2 = 4
Find the equation of the hyperbola, referred to its principal axes as axes of coordinates, in the distance between the foci = 16 and eccentricity = \[\sqrt{2}\].
Find the equation of the hyperbola, referred to its principal axes as axes of coordinates, in the conjugate axis is 7 and passes through the point (3, −2).
Find the equation of the hyperbola whose vertices are (−8, −1) and (16, −1) and focus is (17, −1).
Find the equation of the hyperboala whose focus is at (4, 2), centre at (6, 2) and e = 2.
If P is any point on the hyperbola whose axis are equal, prove that SP. S'P = CP2.
Find the equation of the hyperbola satisfying the given condition :
vertices (0, ± 3), foci (0, ± 5)
find the equation of the hyperbola satisfying the given condition:
vertices (± 7, 0), \[e = \frac{4}{3}\]
Show that the set of all points such that the difference of their distances from (4, 0) and (− 4,0) is always equal to 2 represents a hyperbola.
Write the equation of the hyperbola whose vertices are (± 3, 0) and foci at (± 5, 0).
Equation of the hyperbola whose vertices are (± 3, 0) and foci at (± 5, 0), is
The difference of the focal distances of any point on the hyperbola is equal to
The foci of the hyperbola 2x2 − 3y2 = 5 are
Find the equation of the hyperbola with eccentricity `3/2` and foci at (± 2, 0).
Show that the set of all points such that the difference of their distances from (4, 0) and (– 4, 0) is always equal to 2 represent a hyperbola.
Find the equation of the hyperbola with vertices (± 5, 0), foci (± 7, 0)
Find the equation of the hyperbola with vertices (0, ± 7), e = `4/3`
The locus of the point of intersection of lines `sqrt(3)x - y - 4sqrt(3)k` = 0 and `sqrt(3)kx + ky - 4sqrt(3)` = 0 for different value of k is a hyperbola whose eccentricity is 2.
The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half of the distance between the foci is ______.
