# Find the length of transverse axis, length of conjugate axis, the eccentricity, the co-ordinates of foci, equations of directrices and the length of latus rectum of the hyperbola: x225-y216 = – 1 - Mathematics and Statistics

Sum

Find the length of transverse axis, length of conjugate axis, the eccentricity, the co-ordinates of foci, equations of directrices and the length of latus rectum of the hyperbola:

x^2/25 - y^2/16 = – 1

#### Solution

Given equation of the hyperbola is x^2/25 - y^2/16 = – 1.

∴ y^2/16 - x^2/25 = 1

Comparing this equation with

y^2/"b"^2 - x^2/"a"^2 = 1, we get

b2 = 16 and a2 = 25

∴ b = 4 and a = 5

Length of transverse axis = 2b = 2(4) = 8

Length of conjugate axis = 2a = 2(5) = 10

Co-ordinates of vertices are

B(0, b) and B'(0, – b)

i.e., B(0, 4) and B'(0, – 4)

We know that

e = sqrt("b"^2 + "a"^2)/"b"

= sqrt(16 + 25)/4

= sqrt(41)/4

Co-ordinates of foci are S(0, be) and S'(0, –be),

i.e., "S"(0, 4(sqrt(41)/4)) and "S""'"(0, - 4(sqrt(41)/4)),

i.e., "S"(0, sqrt(41)) and "S""'"(0, -sqrt(41))

Equations of the directrices are y = ± "b"/"e".

∴ y = ± 4/(sqrt(41)/4

∴ y = ± 16/sqrt(41)

Length of latus-rectum = (2"a"^2)/"b"

= (2(25))/4

= 25/2.

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