Thread: Finding area of square in an ellipse

What is the area of a square inscribed in an ellipse 16x² + 9y² = 144?

This is a harder one. I can only work up to the value of a and b but do not know how to find the area of square.

16x²/144 + 9y²/144 = 1

x²/9 + y²/16 = 1

So I know a²=9, therefore a=3 and b²=16 and b=4.

Because a < b, I know this is a vertical ellipse.

The next step is to find the area of a square inscribed in the ellipse. But how?

Thanks.

Originally Posted by shenton

What is the area of a square inscribed in an ellipse 16x² + 9y² = 144?

This is a harder one. I can only work up to the value of a and b but do not know how to find the area of square.

16x²/144 + 9y²/144 = 1

x²/9 + y²/16 = 1

So I know a²=9, therefore a=3 and b²=16 and b=4.

Because a < b, I know this is a vertical ellipse.

The next step is to find the area of a square inscribed in the ellipse. But how?

Thanks.

The corners of the square will be 4 points (x,y) which satisfy the equation of the ellipse and also x^2 = y^2 since it's a square and thus the sides have equal length.

I'll explain how you actually graph the ellipse becaue the previous poster didn't.

You have the find the foci

c^2 = a^2 - b^2

a= 4 and b=3

c^2 = 16 - 9
c= (sqrt)7

the foci are (0, +/- (sqrt)7)

the vertices are (0, +/- 4)

So just graph that and then draw a square inside of the ellipse. Make sure your graph is accurate and then count the sides of the square and multiply the base times the height to get the area.

Originally Posted by shenton

What is the area of a square inscribed in an ellipse 16x² + 9y² = 144?...

The next step is to find the area of a square inscribed in the ellipse. But how?

Thanks.

Hello, shenton,

the vertices of the square lay on the straight lines y = x or y = -x. These lines are the diagonals of the square. (see attachment).

You only have to calculate the intersects of ellipse and straight line:

x²/9 + y²/16 = 1 and y = x ===> x²/9 + x²/16 = 1 ===> 25/144 x² = 1 ===> x = ± 12/5

Therefore the 4 vertices of the square are: (12/5, 12/5), (12/5, -12/5),(-12/5, 12/5), (-12/5, -12/5)

The length of one side of the square is: 12/5 - (-12/5) = 24/5

Therefore the area of the square is (24/5)² = 576/25 = 23.04

EB

Hello, shenton!

This is JakeD's solution ... with a diagram.

What is the area of a square inscribed in an ellipse 16x² + 9y² = 144?

Code:

                      |
                     ***
           B    *     |     *    A
        (-p,q)*-------+-------*(p,q)
             *:       |       :*
              :       |       :
            * :       |       : *
      - - - * : - - - + - - - : * - - -
            * :       |       : *
              :       |       :
             *:       |       :*
              *-------+-------*(p,-q)
                *     |     *    C
                     ***
                      |

We want points A, B, C on the ellipse such that: .AB = AC.
. . That is: .2p = 2q . → . p = q

. . . . . . . . . . . . . . . . . . . . . . . . . . . . _____
Solve the equation for y: . y .= .±(4/3)√9 - x²
. . . . . . . . . . . . - - . . . . . . . . . . . ._____
Then p = q becomes: .p .= .±(4/3)√9 - p²

. . and we have: .9p² .= .16(9 - p²) . → . x² .= .144/25

Hence: .p = 12/5 = 2.4 .and the side of the square is 4.8 units.


Therefore, the area of the square is: .4.8² = 23.04 units².