Diamonds are an Ellipse's Best Friend

Geometry Level 1

The figure above shows two ellipses whose major axes are perpendicular to each other. Each ellipse passes through the other ellipse’s foci, which form the vertices of a square. If the shaded square encloses an area of 16, then what is the area enclosed by one of the ellipses?

8\pi\sqrt{2}

16\pi

8\pi

16\pi\sqrt{2}

1 solution

Andrew Ellinor
Oct 1, 2015

The length of the side of the square is 4, so the diagonal of the square is $4\sqrt{2}$ , which also happens to be the distance between the foci and the length of the minor axis. This means the semi-minor axis has length $b = 2\sqrt{2}$ and the distance from the center to one of the foci is $c = 2\sqrt{2}$ . Using the formula that relates these values to the length of the semi-major axis $a^2 - b^2 = c^2 \longrightarrow a^2 - (2\sqrt{2})^2 = (2\sqrt{2})^2 \longrightarrow a = 4$ .

This means the area of each ellipse is $(4)(2\sqrt{2})\pi = 8\pi\sqrt{2}$ .

Or you can just refer to this .

With $a = 4$ and because it is a square, then $b = c$ , by Pythagoras Theorem, $b = \dfrac4{\sqrt2}$ . Then the area is simply $\pi ab = 8\sqrt2\pi$ .

Pi Han Goh - 5 years, 8 months ago

Yesss.......!

Nihar Mahajan - 5 years, 8 months ago

Diamonds are an Ellipse's Best Friend

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