FIND THE EQUATION OF THE HYPERBOLA WITH THE GIVEN INFORMATION

About "Find the Equation of the Hyperbola with the Given Information"

Find the Equation of the Hyperbola with the Given Information :

Here we are going to see some practice questions on find the equation of the hyperbola with the given information.

Find the Equation of the hyperbola with the Given Information - Practice questions

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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