Question -
Answer -
(i) 9x2 – 16y2 = 144
Given:
The equation => 9x2 – 16y2 =144
The equation can be expressed as:
The obtained equation is of the form
Where, a2 = 16, b2 = 9 i.e.,a = 4 and b = 3
Eccentricity is given by:
Foci: The coordinates of the foci are (0, ±be)
(0, ±be) = (0, ±4(5/4))
= (0, ±5)
The equation of directrices is given as:
5x ∓ 16 = 0
The length of latus-rectum is given as:
2b2/a
= 2(9)/4
= 9/2
(ii) 16x2 – 9y2 = -144
Given:
The equation => 16x2 – 9y2 =-144
The equation can be expressed as:
The obtained equation is of the form
Where, a2 = 9, b2 = 16 i.e.,a = 3 and b = 4
Eccentricity is given by:
Foci: The coordinates of the foci are (0, ±be)
(0, ±be) = (0, ±4(5/4))
= (0, ±5)
The equation of directrices is given as:
The length of latus-rectum is given as:
2a2/b
= 2(9)/4
= 9/2
(iii) 4x2 – 3y2 = 36
Given:
The equation => 4x2 – 3y2 =36
The equation can be expressed as:
The obtained equation is of the form
Where, a2 = 9, b2 = 12 i.e.,a = 3 and b = √12
Eccentricity is given by:
Foci: The coordinates of the foci are (±ae, 0)
(±ae, 0) = (±√21,0)
The equation of directrices is given as:
The length of latus-rectum is given as:
2b2/a
= 2(12)/3
= 24/3
= 8
(iv) 3x2 – y2 = 4
Given:
The equation => 3x2 – y2 =4
The equation can be expressed as:
The obtained equation is of the form
Where, a = 2/√3 and b = 2
Eccentricity is given by:
Foci: The coordinates of the foci are (±ae, 0)
(±ae, 0) = ±(2/√3)(2) = ±4/√3
(±ae, 0) = (±4/√3, 0)
The equation of directrices is given as:
The length of latus-rectum is given as:
2b2/a
= 2(4)/[2/√3]
= 4√3
(v) 2x2 – 3y2 = 5
Given:
The equation => 2x2 – 3y2 =5
The equation can be expressed as:
The obtained equation is of the form
Where, a = √5/√2 and b = √5/√3
Eccentricity is given by:
Foci: The coordinates of the foci are (±ae, 0)
(±ae, 0) = (±5/√6, 0)
The equation of directrices is given as:
The length of latus-rectum is given as:
2b2/a
