Advertisements
Advertisements
प्रश्न
Find the equation of the hyperbola with eccentricity `3/2` and foci at (± 2, 0).
उत्तर
Given that e = `3/2` and foci at (± 2, 0)
We know that foci = (± ae, 0)
∴ ae = 2
⇒ `a xx 3/2` = 2
⇒ `a = 4/3`
⇒ `a^2 = 16/9`
We know that b2 = a2(e2 – 1)
⇒ `b^2 = 16/9(9/4 - 1)`
= `16/9 xx 5/4`
= `20/9`
So, the equation of the hyperbola is `x^2/(16/9) - y^2/(20/9)` = 1
⇒ `(9x^2)/16 - (9y^2)/20` = 1
⇒ `x^2/4 - y^2/5 = 4/9`
Hence, the required equation is `x^2/4 - y^2/5 = 4/9`.
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:
Foci (0, ±13), the conjugate axis is of length 24.
Find the equation of the hyperbola satisfying the given conditions:
Foci `(0, +- sqrt10)`, passing through (2, 3)
Find the equation of the hyperbola whose focus is (1, 1), directrix is 3x + 4y + 8 = 0 and eccentricity = 2 .
Find the equation of the hyperbola whose focus is (2, −1), directrix is 2x + 3y = 1 and eccentricity = 2 .
Find the equation of the hyperbola whose focus is (2, 2), directrix is x + y = 9 and eccentricity = 2.
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 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 hyperbola whose vertices are at (0 ± 7) and foci at \[\left( 0, \pm \frac{28}{3} \right)\] .
Find the equation of the hyperbola whose vertices are at (± 6, 0) and one of the directrices is x = 4.
Find the equation of the hyperbola whose foci at (± 2, 0) and eccentricity is 3/2.
Find the equation of the hyperboala whose focus is at (5, 2), vertex at (4, 2) and centre at (3, 2).
Find the equation of the hyperboala whose focus is at (4, 2), centre at (6, 2) and e = 2.
Find the equation of the hyperbola satisfying the given condition:
foci (0, ± \[\sqrt{10}\], passing through (2, 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.
The difference of the focal distances of any point on the hyperbola is equal to
The foci of the hyperbola 2x2 − 3y2 = 5 are
The equation of the hyperbola whose centre is (6, 2) one focus is (4, 2) and of eccentricity 2 is
Find the equation of the hyperbola with vertices at (0, ± 6) and e = `5/3`. Find its foci.
The eccentricity of the hyperbola `x^2/a^2 - y^2/b^2` = 1 which passes through the points (3, 0) and `(3 sqrt(2), 2)` is ______.
Find the equation of the hyperbola with vertices (0, ± 7), e = `4/3`
Find the equation of the hyperbola with foci `(0, +- sqrt(10))`, passing through (2, 3)
Equation of the hyperbola with eccentricty `3/2` and foci at (± 2, 0) is ______.
