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

Answer 1

(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