Find the equation of the hyperbola satisfying the given conditions: Foci (±3√5,0) the latus rectum is of length 8

Solution:

The problem statement can be summarized in the diagram below:

PF₁ - PF₂ = 2a---------->(1)

F₁F₂ = 2c = 2(3√5) = 6√5---------->(2)

PF₂ = 4 (half the length of the latus rectum which is 8)

From equation (1) we know that PF₁ - PF₂ = 2a

Since △PF₂F₁ is a right angled triangle we can write

(PF₁)² = (PF₂)² + (F₁F₂)²---------->(3)

Since the foci  F₁ and F₂ are (-3√5, 0) and (3√5, 0) respectively we can state that

Length (F₁F₂) = 6√5

Substituting the value of F₁F₂ in (3) we obtain

(PF₁)² = (4)²  +  ( 6√5)²

=  16  +  180

=   196

Therefore

PF₁ = √196 = 14

Now the property of hyperbola states:

PF₁ - PF₂ = 2a

14 - 4 = 2a

2a = 10

a = 5

If the foci is on the x-axis in this case then the standard form of equation for a hyperbola will be 

x² / a² - y² / b²    =  1

Where b² = c²  - a²

 b² = (3√5)²  -  (5)²  = 45 - 25 = 20

Hence the equation of the hyperbola can be written as:

x² / 25 -  y² / 20    =  1

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Find the equation of the hyperbola satisfying the given conditions: Foci (±3√5,0) the latus rectum is of length 8

Summary:

The equation of the hyperbola satisfying the given conditions i.e.  Foci (±3√5,0) and the latus rectum  of length 8 is    x² / 25 -  y² / 20    =  1