Question

if the latusrectum of a hyperbola through one focus subtends 60 degrees angle at the other focus then its eccentricity is

Answer 1

Answer:

e = √3.

Step-by-step explanation:

Hi,

Let the equation of the hyperbola be x²/a² - y²/b² = 1

So, the foci are S(ae,0)  and S'(-ae,0).

End points of the lactus rectum through S are given A(ae, b²/a) and

B(ae, -b²/a).

Now given that AB subtend an angle of 60° at S'

=> Angle between S'A with x -axis will be is 30° (due to symmetry)

Slope of S'A , m₁= b²/(2a²e) = tan30°

=> b²/a² = 2e/√3

But  b²/a² = e² - 1

=> e² - 1 = 2e/√3

=> e² -2e/√3 - 1 = 0

=> e = (2/√3 ±√4/3+4)/2

=> Since 'e' cannot take negative values

e = 3/√3 = √3

Hope , it helped !