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The equation of the directrix of a hyperbola is x – y + 3 = 0. Its focus is (-1, 1) and eccentricity 3. Find the equation of the hyperbola.
Find the equation of the hyperbola whose
focus is (0, 3), directrix is x + y – 1 = 0 and eccentricity = 2
focus is (1, 1), directrix is 3x + 4y + 8 = 0 and eccentricity = 2
focus is (1, 1) directrix is 2x + y = 1 and eccentricity =
focus is (2, -1), directrix is 2x + 3y = 1 and eccentricity = 2
focus is (a, 0), directrix is 2x + 3y = 1 and eccentricity = 2
focus is (2, 2), directrix is x + y = 9 and eccentricity = 2
Find the eccentricity, coordinates of the foci, equations of directrices and length of the latus-rectum of the hyperbola.
9x2 – 16y2 = 144
16x2 – 9y2 = -144
4x2 – 3y2 = 36
3x2 – y2 = 4
2x2 – 3y2 = 5
Find the axes, eccentricity, latus-rectum and the coordinates of the foci of the hyperbola 25x2 – 36y2 = 225.
Find the centre, eccentricity, foci and directions of the hyperbola
16x2 – 9y2 + 32x + 36y – 164 = 0
x2 – y2 + 4x = 0
x2 – 3y2 – 2x = 8
Find the equation of the hyperbola, referred to its principal axes as axes of coordinates, in the following cases:
the distance between the foci = 16 and eccentricity =
conjugate axis is 5 and the distance between foci = 13
conjugate axis is 7 and passes through the point (3, -2).
foci are (6, 4) and (-4, 4) and eccentricity is 2.
vertices are (-8, -1) and (16, -1) and focus is (17, -1)
foci are (4, 2) and (8, 2) and eccentricity is 2.
vertices are at (0. ± 7) and foci at
vertices are at (±6, 0) and one of the directrices is x = 4.
foci at (± 2, 0) and eccentricity is 3/2.
Find the eccentricity of the hyperbola, the length of whose conjugate axis is of the length of the transverse axis.
the focus is at (5, 2), vertices at (4, 2) and (2, 2) and centre at (3, 2)
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
In each of the following find the equations of the hyperbola satisfying the given conditions
vertices (± 2, 0), foci (± 3, 0)
vertices (0, ± 5), foci (0, ± 8)
vertices (0, ± 3), foci (0, ± 5)
foci (±5, 0), transverse axis = 8
foci (0, ±13), conjugate axis = 24
foci , the latus-rectum = 8
foci (±4, 0), the latus-rectum = 12
vertices (0, ± 6),
foci , passing through (2, 3)
foci (0, ± 12), latus-rectum = 36.
If the distance between the foci of a hyperbola is 16 and its eccentricity is, then obtain its equation.
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 eccentricity of the hyperbola 9x2 – 16y2 = 144.
Write the eccentricity of the hyperbola whose latus-rectum is half of its transverse axis.
Write the coordinates of the foci of the hyperbola 9x2 – 16y2 = 144.
Write the equation of the hyperbola of eccentricity, if it is known that the distance between its foci is 16.
If the foci of the ellipse and the hyperbola coincide, write the value of b2
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.
Write the distance between the directrices of the hyperbola
x = 8 sec θ, y = 8 tan θ.
Write the equation of the hyperbola whose vertices are (±3, 0) and foci at (±5, 0).
If e1 and e2 are respectively the eccentricities of the ellipse and the hyperbola , then write the value of 2e12 + e22.
Equation of the hyperbola whose vertices are (± 3, 0) and foci at (± 5, 0), is
If e1 and e2 are respectively the eccentricities of the ellipse and the hyperbola , then the relation between e1 and e2 is
The distance between the directrices of the hyperbola x = 8 sec θ, y = 8 tan θ, is
The equation of the conic with focus at (1, -1) directrix along x – y + 1= 0 and eccentricity is
The eccentricity of the conic 9x2 – 16y2 = 144 is
A point moves in a plane so that its distances PA and PB from two fixed points A and B in the plane satisfy the relation PA – PB = k(k ≠ 0), then the locus of P 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 difference of the focal distances of any point on the hyperbola is equal to
the foci of the hyperbola 9x2 – 16y2 = 144 are
The distance between the foci of a hyperbola is 16 and its eccentricity is , then equation of the hyperbola is
If e1 is the eccentricity of the conic 9x2 + 4y2 = 36 and e2 is the eccentricity of the conic 9x2 – 4y2 = 36, then
If the eccentricity of the hyperbola x2 – y2 sec2∝ = 5 is times the eccentricity of the ellipse x2 sec2∝ +y2 = 25, then ∝ =
The equation of the hyperbola whose foci are (6, 4) and (-4, 4) and eccentricity 2, is
The length of the straight line x – 3y = 1 intercepted by the hyperbola x2 – 4y2 = 1 is
The latus-rectum of the hyperbola 16x2 – 9y2 = 144 is
The foci of the hyperbola 2x2 – 3y2 = 5 are
The eccentricity the hyperbola is
The equation of the hyperbola whose centre is (6, 2) one focus is (4, 2) and of eccentricity 2 is
The locus of the point of intersection of the lines and is a hyperbola of eccentricity
