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\[\large \frac{x^2}{16}+\frac{y^2}{9}=1\]

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 $\triangle OAB$?

 Jan 17, 2021
 #1
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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

 

 

cool cool cool

 Jan 17, 2021

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