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Question

Find the eccentricity, coordinates of the foci, equations of directrices and length of the latus-rectum of the hyperbola
(i) 9x2 − 16y2 = 144
(ii) 16x2 − 9y2 = −144
(iii) 4x2 − 3y2 = 36
(iv) 3x2 − y2 = 4
(v) 2x2 − 3y2 = 5.

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Solution

(i) Equation of the hyperbola:
9x2-16y2=144
This can be rewritten in the following way:
x216-y29=1
This is the standard equation of a hyperbola, where a2=16 and b2=9.

b2=a2(e2-1)9=16(e2-1)e2-1=916e2=2516e=54
Coordinates of the foci are given by ±ae,0, i.e. ±5,0.
Equation of directrices:
x=±ae
x=±4545x±16=0
Length of the latus rectum of the hyperbola is 2b2a.
Length of the latus rectum = 2×94=92

(ii) Equation of the hyperbola:
16x2-9y2=-144
This can be rewritten in the following way:
x29-y216=-1
This is the standard equation of a hyperbola, where a2=9 and b2=16.

a2=b2(e2-1)9=16(e2-1)e2-1=916e2=2516e=54
Coordinates of foci are given by 0,±ae, i.e. 0,±5.
Equation of the directrices:
y=±ae
y=±4545y±16=0
Length of the latus rectum of the hyperbola = 2a2b
Length of the latus rectum = 2×94=92

(iii) Equation of the hyperbola:
4x2-3y2=36
This can be rewritten in the following way:
4x236-3y236=1x29-y212=1
This is the standard equation of a hyperbola, where a2=9 and b2=12.

b2=a2(e2-1)12=9(e2-1)e2-1=43e2=73e=73
Coordinates of the foci are given by ±ae,0, i.e. ±21,0.
Equation of the directrices:
x=±ae
x=±3737x±33=0
Length of the latus rectum of the hyperbola is 2b2a.
2×123=8

(iv) Equation of the hyperbola:
3x2-y2=4
This can be rewritten in the following way:
3x24-y24=1x243-y24=1
This is the standard equation of a hyperbola, where a2=43 and b2=4.

b2=a2(e2-1)4=43(e2-1)e2-1=3e2=4e=2
Coordinates of the foci are given by ±ae,0, i.e. ±433,0.
Equation of the directrices:
x=±ae
x=±4323x±1=0
Length of the latus rectum of the hyperbola = 2b2a
2×443=43

(v) Equation of the hyperbola:
2x2-3y2=5
This can be rewritten in the following manner:
2x25-3y25=1x252-y253=1
This is the standard equation of a hyperbola, where a2=52 and b2=53.

b2=a2(e2-1)53=52(e2-1)e2-1=23e2=53e=53
Coordinates of the foci are given by ±ae,0, i.e. ±566,0.
Equation of the directrices:
x=±ae
x=±5253x=±322x±3 = 0
Length of the latus rectum of the hyperbola is 2b2a.

2×5352=10325

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