x^2/16+y^2/81

Question

Identify the conic as a circle or an ellipse. Then find the center, radius, vertices, foci, and eccentricity of the conic (if applicable), and sketch its graph. Help, the only thing I know is that its an ellipse.

Answer 1

Assuming you meant to say

 

x²/16 + y²/81 = 1,

 

you have an ellipse with semi-major axis a = √81 = 9 along the y-axis, and semi-minor axis b = √16 = 4 along the x-axis.  

 

The center is at (0,0)

 

Eccentricity:

 

e = √[1 - (b/a)²] = √[1 - (4/9)²] ≅ 0.8958 < 1

 

Foci:

 

c = ae = 9(0.8958) = 8.0622

 

Foci are at (0,±8.0622)

 

This forum is not set up for graphing, but knowing a and b, and knowing the semi-major axis is along the y-axis, and that the center is at the origin, you should be able to make a sketch.

Answer 2

Hi Ashely,

 

x^2/16 + y^2/81 should be equal to something for it to define a conic section like an ellipse. Is it equal to 1?