Kelton Bailey

2022-10-01

Finding the foci and vertices of an ellipse.
How would you find the foci and vertices of the following ellipse: $\frac{2{x}^{2}}{15}+\frac{8{y}^{2}}{45}-\frac{2\sqrt{3}}{45}xy=1?$

Matteo Estes

Step 1
Notice first of all that your ellipse is centered at $O=\left(0,0\right)$ (because if (x,y) belongs to the ellipse, then also (-x, -y) belongs to it).
To find the axes of the ellipse, notice that if $P=\left(x,y\right)$ is a vertex, then the tangent at P is perpendicular to PO, that is ${y}^{\prime }\left(y/x\right)=-1$. You can compute y′ by differentiating the equation of the ellipse:
$\frac{4}{15}x+\frac{16}{45}y{y}^{\prime }-\frac{2\sqrt{3}}{45}y-\frac{2\sqrt{3}}{45}x{y}^{\prime }=0,$
whence: ${y}^{\prime }=\frac{2\sqrt{3}y-12x}{16y-2\sqrt{3}x}.$
Step 2
The above condition ${y}^{\prime }\left(y/x\right)=-1$ implies then that the coordinates of a vertex are related by:
$\frac{y}{x}=\frac{-1±2}{\sqrt{3}}.$
Plugging that into the ellipse equation you can get the coordinates of the vertices, and then of course those of the foci.

seguitzla

Step 1
Consider a rotation: $x=X\mathrm{cos}\alpha -Y\mathrm{sin}\alpha ,\phantom{\rule{2em}{0ex}}y=X\mathrm{sin}\alpha +Y\mathrm{cos}\alpha$
Then your ellipse becomes $6\left(X\mathrm{cos}\alpha -Y\mathrm{sin}\alpha {\right)}^{2}+8\left(X\mathrm{sin}\alpha +Y\mathrm{cos}\alpha {\right)}^{2}-2\sqrt{3}\left(X\mathrm{cos}\alpha -Y\mathrm{sin}\alpha \right)\left(X\mathrm{sin}\alpha +Y\mathrm{cos}\alpha \right)=45$
The term in XY has coefficient $4\mathrm{cos}\alpha \mathrm{sin}\alpha -2\sqrt{3}{\mathrm{cos}}^{2}\alpha +2\sqrt{3}{\mathrm{sin}}^{2}\alpha$ which you want to be vanishing; dividing by ${\mathrm{cos}}^{2}\alpha$ we get $2\sqrt{3}{\mathrm{tan}}^{2}\alpha +4\mathrm{tan}\alpha -2\sqrt{3}=0$ so $\mathrm{tan}\alpha =\frac{-2+4}{2\sqrt{3}}=\frac{1}{\sqrt{3}}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\mathrm{tan}\alpha =\frac{-2-4}{2\sqrt{3}}=-\sqrt{3}$
Step 2
So we can take $\alpha =\pi /6$ and the equation of the ellipse becomes
${X}^{2}\left(6{\mathrm{cos}}^{2}\alpha +8{\mathrm{sin}}^{2}\alpha -2\sqrt{3}\mathrm{cos}\alpha \mathrm{sin}\alpha \right)+{Y}^{2}\left(6{\mathrm{sin}}^{2}\alpha +8{\mathrm{cos}}^{2}\alpha +2\sqrt{3}\mathrm{cos}\alpha \mathrm{sin}\alpha \right)=45$
that is $5{X}^{2}+9{Y}^{2}=45$ or $\frac{{X}^{2}}{9}+\frac{{Y}^{2}}{5}=1$.
Find foci and vertices, then use the inverse rotation.

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