Question

Find the vertices and foci of the ellipse. x2 + 3y2 + 4x − 12y + 13 = 0

Answer 1

Hey there hope I can help!

NOTE: Please look at the images as they are useful tips!
$\left(h+c,\:k\right),\:\left(h-c,\:k\right)$

$\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1\:\mathrm{is\:the\:ellipse\:standard\:equation}$
$\mathrm{with\:center}\:\left(h,\:k\right)\:\mathrm{and\:}a,\:b\mathrm{\:are\:the\:semi-major\:and\:semi-minor\:axes}$

$x^2+3y^2+4x-12y+13=0 \ \textgreater \ \mathrm{Subtract\:}13\mathrm{\:from\:both\:sides}$
$x^2+4x-12y+3y^2=-13$

$\mathrm{Factor\:out\:coefficient\:of\:square\:terms} \ \textgreater \ \left(x^2+4x\right)+3\left(y^2-4y\right)=-13$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}3$
$\frac{1}{3}\left(x^2+4x\right)+\left(y^2-4y\right)=-\frac{13}{3}$

$\mathrm{Convert}\:x\:\mathrm{to\:square\:form}$
$\frac{1}{3}\left(x^2+4x+4\right)+\left(y^2-4y\right)=-\frac{13}{3}+\frac{1}{3}\left(4\right)$

$\mathrm{Convert\:to\:square\:form}$
$\frac{1}{3}\left(x+2\right)^2+\left(y^2-4y\right)=-\frac{13}{3}+\frac{1}{3}\left(4\right)$

$\mathrm{Convert}\:y\:\mathrm{to\:square\:form}$
$\frac{1}{3}\left(x+2\right)^2+\left(y^2-4y+4\right)=-\frac{13}{3}+\frac{1}{3}\left(4\right)+4$

$\mathrm{Convert\:to\:square\:form}$
$\frac{1}{3}\left(x+2\right)^2+\left(y-2\right)^2=-\frac{13}{3}+\frac{1}{3}\left(4\right)+4$

$\mathrm{Refine\:}-\frac{13}{3}+\frac{1}{3}\left(4\right)+4 \ \textgreater \ \frac{1}{3}\left(x+2\right)^2+\left(y-2\right)^2=1$

Refine again
$\frac{\left(x+2\right)^2}{3}+\frac{\left(y-2\right)^2}{1}=1$

Rewrite it in the ellipse standard form
$\frac{\left(x-\left(-2\right)\right)^2}{\left(\sqrt{3}\right)^2}+\frac{\left(y-2\right)^2}{1^2}=1$

$\mathrm{Therefore\:ellipse\:properties\:are:}\left(h,\:k\right)=\left(-2,\:2\right),\:a=\sqrt{3},\:b=1$
$\left(-2+c,\:2\right),\:\left(-2-c,\:2\right)$

Now we want to compute c!

$c =\sqrt{\left(\sqrt{3}\right)^2-1^2} \ \textgreater \ \mathrm{Apply\:rule}\:1^a=1 \ \textgreater \ 1^2=1 \ \textgreater \ \sqrt{\left(\sqrt{3}\right)^2-1}$

$\left(\sqrt{3}\right)^2 \ \textgreater \ \sqrt{a}=a^{\frac{1}{2}} \ \textgreater \ \left(3^{\frac{1}{2}}\right)^2 \ \textgreater \ \mathrm{Apply\:exponent\:rule}: \left(a^b\right)^c=a^{bc} \ \textgreater \ 3^{\frac{1}{2}\cdot \:2}$

$\frac{1}{2}\cdot \:2 \ \textgreater \ \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \ \frac{1\cdot \:2}{2} \ \textgreater \ \mathrm{Apply\:rule}\:1\cdot \:a=a$
$\frac{2}{2} \ \textgreater \ \mathrm{Cancel\:the\:common\:factor:}\:2 \ \textgreater \ 1 \ \textgreater \ 3 \ \textgreater \ \sqrt{3-1} \ \textgreater \ \sqrt{2}$

Therefore the foci is $\left(-2+\sqrt{2},\:2\right),\:\left(-2-\sqrt{2},\:2\right)$

$\mathrm{The\:vertices\:are\:the\:two\:points\:on\:the\:ellipse\:that\:intersect\:the\:major\:axis}$
$\left(h+a,\:k\right),\:\left(h-a,\:k\right)$

$x^2+3y^2+4x-12y+13=0 \ \textgreater \ \mathrm{Subtract\:}13\mathrm{\:from\:both\:sides}$
$x^2+4x-12y+3y^2=-13$

$\mathrm{Factor\:out\:coefficient\:of\:square\:terms} \ \textgreater \ \left(x^2+4x\right)+3\left(y^2-4y\right)=-13$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}3:\frac{1}{3}\left(x^2+4x\right)+\left(y^2-4y\right)=-\frac{13}{3}$

$\mathrm{Convert}\:x\:\mathrm{to\:square\:form}$
$\frac{1}{3}\left(x^2+4x+4\right)+\left(y^2-4y\right)=-\frac{13}{3}+\frac{1}{3}\left(4\right)$

$\mathrm{Convert\:to\:square\:form}$
$\frac{1}{3}\left(x+2\right)^2+\left(y^2-4y\right)=-\frac{13}{3}+\frac{1}{3}\left(4\right)$

$\mathrm{Convert}\:y\:\mathrm{to\:square\:form}$
$\frac{1}{3}\left(x+2\right)^2+\left(y^2-4y+4\right)=-\frac{13}{3}+\frac{1}{3}\left(4\right)+4$

$\mathrm{Convert\:to\:square\:form}$
$\frac{1}{3}\left(x+2\right)^2+\left(y-2\right)^2=-\frac{13}{3}+\frac{1}{3}\left(4\right)+4$

$\mathrm{Refine\:}-\frac{13}{3}+\frac{1}{3}\left(4\right)+4 \ \textgreater \ \frac{1}{3}\left(x+2\right)^2+\left(y-2\right)^2=1 \ \textgreater \ Refine$
$\frac{\left(x+2\right)^2}{3}+\frac{\left(y-2\right)^2}{1}=1$

Rewrite in the standard form again
$\frac{\left(x-\left(-2\right)\right)^2}{\left(\sqrt{3}\right)^2}+\frac{\left(y-2\right)^2} {1^2}=1$

$\mathrm{Therefore\:ellipse\:properties\:are:} \left(h,\:k\right)=\left(-2,\:2\right),\:a=\sqrt{3},\:b=1$

Which means the vertices are $\left(-2+\sqrt{3},\:2\right),\:\left(-2-\sqrt{3},\:2\right)$

Hope this helps!