\[\large \frac{x^2}{16}+\frac{y^2}{9}=1\]
The equation above is an ellipse with center O. A chord $AB=2 \sqrt{7}$ is parallel to the major axis. What is the area of $\triangle OAB$?
+128408 The center of the ellipse is (0,0) = O
1/2 of the chord length will be the x value associated with one of the other vertexes on the positive side of the y axis ......the other vertex will have an x coordinate of - sqrt(7)
So.....to find the y value we have
(sqrt 7)^2 / 16 + y^2/9 = 1
7/16 + y^2/9 = 1
y^2/9 = 1 - 7/16
y^2/9 = 9/16 multiply both sides by 9
y^2 = 81/16 take the positive root
y= 9/4 = the height of the triangle
The area of the triangle = (1/2) chord length * height = sqrt (7) *(9/4) = 2.25 sqrt (7) units^2
