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.