# Finding area of square in an ellipse

• Mar 11th 2007, 09:48 PM
shenton
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.
• Mar 11th 2007, 10:17 PM
JakeD
Quote:

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.
• Mar 11th 2007, 10:25 PM
zachb
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.
• Mar 11th 2007, 11:43 PM
earboth
Quote:

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

• Mar 12th 2007, 06:33 AM
Soroban
Hello, shenton!

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

Quote:

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) . . .= .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.