Question

An ellipse has equation 121x^2+9y^2 = 1089. Find its foci. (i) (0, ±4√7) (ii) (±4√7, 0) (iii) (0, ±11) (iv) (±11,0) (v) (±3,0) (vi) (0, ±3) (vii) not listed

Answer 1

In conclusion, the foci of the given ellipse are located at (±4√7, 0), which is option (ii) listed in the question.

Step-by-step explanation:

The equation of the ellipse appears to be in standard form: a^2x^2 + b^2y^2 = c^2. To find the foci of the ellipse, we first need to determine the semi-major axis (a) and the semi-minor axis (b). From the equation 121x^2 + 9y^2 = 1089, we can see that a^2 = 121 (thus, a=11) and b^2 = 9 (thus, b=3).

The foci of an ellipse are located along its major axis at a distance of ae away from the center, where e is the eccentricity of the ellipse, and e = sqrt(1 - (b^2/a^2)). Substituting the values for b and a, we get e = sqrt(1 - (3^2/11^2)) = sqrt(1 - 9/121) = sqrt(112/121) = 4√7/11.

Since the x term has a larger coefficient in the equation, we can see that the ellipse is oriented along the x-axis. Thus, the foci will be located at (±ae, 0) = (±4√7, 0), so the option (ii) can be the correct answer.

Learn more about Foci of an Ellipse