find the ecentricity and coordinates of foci of the ellipse 16x2 + 9y2= 144
x^2/9 + y^2/16 = 1
a=3, b=4, so c=√7
e = c/a
To find the eccentricity and coordinates of the foci of the ellipse 16x^2 + 9y^2 = 144, we can start by putting the equation in the standard form for an ellipse:
x^2/a^2 + y^2/b^2 = 1
Where a is the semi-major axis length and b is the semi-minor axis length.
Dividing the given equation by 144, we get:
x^2/9 + y^2/16 = 1
Comparing this with the standard form, we can see that a^2 = 9 and b^2 = 16.
To find the semi-major axis length (a) and semi-minor axis length (b), we take the square root of a^2 and b^2:
a = √(9) = 3
b = √(16) = 4
Now, we can find the eccentricity (e) using the formula:
e = √(1 - (b^2/a^2))
Substituting the values of a and b:
e = √(1 - (4^2/3^2))
e = √(1 - 16/9) = √(9/9 - 16/9) = √(-7/9)
As the result is a complex number, the eccentricity (e) of the ellipse is an imaginary number.
Finally, the coordinates of the foci can be found using the formula:
c = √(a^2 - b^2)
Substituting the values of a and b we found earlier:
c = √(3^2 - 4^2) = √(9 - 16) = √(-7)
Again, the result is a complex number, so the coordinates of the foci are not real.