What is the equation of ellipse if the major axis is 10 and the distance between focci is 2c=8?
The equation of the ellipse with major axis 2a and minor axis b i given by:
x^2/a^2+y^2/b^2 = 1.
Given major axis = 10.Therefore 2 a= 10 , or a = 10/2 = 5.
The distance between the focii 2c = 8.Therefor 2ae = 8, where e is the eccentricity of the ellipse.
e = 8/2a = 8/10 = 0.8.
Therefore b = a(1- e^2)^(1/2) = 5(1-0.8)^2 = 5*(1-0.64)^(1/2) = 5*(0.36)^(1/2) = 5*0.6) = 3. So b= 3.
So a = 5 and b = 3. Therefore the equation of the ellipse is x^2/a^2+y^2/b^2 = 1. Or x^2/5^2+y^2/3^2 = 0.
We multiply x^2/5^2+y^2/3^2 = 1 by 5^2*3^2 = 225. Then we get the equation of the ellipse : 9x^2+25y^2 = 225.
Therefore the required equation of the ellipse is
x^2/25+y^2/9 = 1 or 9x^2+25y^2 = 225.
We'll write the equation of the ellipse:
x^2/a^2 + y^2/b^2 = 1.
This equation is determined if the values of the major semi-axis,a, and minor semi-axis,b, are known.
The relation between semi-axis and the semi-distance between foci is:
a^2 = b^2 + c^2
Since we know a and c, we'll determine b by subtracting c^2 both sides:
but, 2c=8, so c=4
We know that 2a=10, so a=5
We'll substitute a and c and we'll get:
b= sqrt(5^2 - 4^2)
b = sqrt(25 - 16)
b = sqrt 9
The equation of the ellipse, with a=5 and b=3 is
x^2/25 + y^2/9 = 1