FIND THE EQUATION OF THE HYPERBOLA WITH THE GIVEN INFORMATION

Question 1 :

Find the equation of the hyperbola in each of the cases given below:

(i) foci (± 2, 0) , eccentricity = 3/2.

Solution :

F₁ (2, 0) and F₂ (-2, 0) and e = 3/2

From the given information, we know that the given ellipse is symmetric about x axis.

Midpoint of foci  =  Center of the ellipse

Center  =  (2 + (-2))/2, (0 + 0)/2  =  C (0, 0)

Distance between foci  =  √(x₂ - x₁)² + (y₂ - y₁)²

  =  √(2 + 2)² + (0 - 0)²

  =  √4² + 0²

2ae  =  4

ae  =  2

e = 3/2

a (3/2) =  2

a  =  4/3 ==>  a²  =  16/9

b²  =  a² (e² - 1)

b²  =  (ae)² - a²

b²  =  4 - (16/9)

b²  =  20/9

Equation of hyperbola :

(x²/a²) - (y²/b²)  =  1

(9x²/16) - (9y²/20)  =  1

(ii) Centre (2, 1) , one of the foci (8, 1) and corresponding directrix x = 4 .

Solution :

Distance between center and foci  = √(x₂-x₁)² + (y₂-y₁)²

ae  =  √(8-2)² + (1-1)²

ae  =  √36

ae  =  6 -----(1)

Equation of directrix x = 4

a/e  =  4 -----(2)

(1) x (2) ==>  a²  =  24

b²  =  a² (e² - 1)

b²  =  (ae)² - a²

b²  =  36 - 24

b²  =  12

[(x-h)²/a²] - [(y-k)²/b²]  =  1

[(x-2)²/24] - [(y-1)²/12]  =  1

(iii) passing through (5,−2) and length of the transverse axis along x axis and of length 8 units.

Solution :

Length of transverse axis  =  8

2a  =  8

a  =  4

a²  =  16

(x²/a²) - (y²/b²)  =  1

The hyperbola passes through the point (5, -2).

(5²/4²) - (2²/b²)  =  1

(25/16)  - (4/b²)  =  1

(25/16) - 1  =  (4/b²)

(4/b²)  =  9/16

64/9  =  b²

(x²/16) - (y²/(64/9))  =  1

(x²/16) - (9y²/64)  =  1

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