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fitzger · Dec 5, 2010, 03:19 AM

find the vertices of an equilaterall triangle circumscribed about the ellipse 9x^2 + 16y^2 = 144.

galactus · Dec 5, 2010, 07:58 AM

There can be two such triangles, One with apex up and one with apex down.

The ellipse has semi-major axis length of 4 and semi-minor axis length of 3.

$\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$

Upper half of ellipse has equation:

$y=\frac{3\sqrt{16-x^{2}}}{4}$

$y'=\frac{-3x}{4\sqrt{16-x^{2}}}$

We know this slope must be $-\sqrt{3}$

because the equilateral triangle has angles of 60 degrees, and

$tan(60)=\sqrt{3}$ . But the line on the right has negative slope.

$\frac{-3x}{4\sqrt{16-x^{2}}}=-\sqrt{3}$

$x=\frac{16}{\sqrt{19}}$ .

Thus, the line is tangent to the ellipse at $(\frac{16}{\sqrt{19}}, \;\ \frac{3\sqrt{57}}{19})$

Using $y-y_{1}=m(x-x_{1})$ we can solve for the lower right vertex of the triangle.

$\frac{3\sqrt{16-x^{2}}}{4}+3=-\sqrt{3}(x-a)$ , where a is the x coordinate of this vertex.

Sub in the newly found x coordinate of the tangent point, $x=\frac{16}{\sqrt{19}}$ , and we get:

$\fbox{(\sqrt{3}+\sqrt{19}, \;\ -3)}$

This means the left vertex has coordinates:

$\fbox{(-\sqrt{3}-\sqrt{19}, \;\ -3)}$

The apex has coordinate:

$-3=-\sqrt{3}(\sqrt{3}+\sqrt{19})+b$

$b=\sqrt{57}$

$\fbox{(0, \;\ \sqrt{57})}$

There is also a triangle that can be positioned with apex down as well.