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)`


`(\left(x-h\right)^2)/(a^2)+(\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`

`(1)/(3)\left(x^2+4x\right)+\left(y^2-4y\right)=-(13)/(3)`


`\mathrm{Convert}\:x\:\mathrm{to\:square\:form}`

`(1)/(3)\left(x^2+4x+4\right)+\left(y^2-4y\right)=-(13)/(3)+(1)/(3)\left(4\right)`


`\mathrm{Convert\:to\:square\:form}`

`(1)/(3)\left(x+2\right)^2+\left(y^2-4y\right)=-(13)/(3)+(1)/(3)\left(4\right)`


`\mathrm{Convert}\:y\:\mathrm{to\:square\:form}`

`(1)/(3)\left(x+2\right)^2+\left(y^2-4y+4\right)=-(13)/(3)+(1)/(3)\left(4\right)+4`


`\mathrm{Convert\:to\:square\:form}`

`(1)/(3)\left(x+2\right)^2+\left(y-2\right)^2=-(13)/(3)+(1)/(3)\left(4\right)+4`


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

Refine again

`(\left(x+2\right)^2)/(3)+(\left(y-2\right)^2)/(1)=1`

Rewrite it in the ellipse standard form

`(\left(x-\left(-2\right)\right)^2)/(\left(√(3)\right)^2)+(\left(y-2\right)^2)/(1^2)=1`


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

`\left(-2+c,\:2\right),\:\left(-2-c,\:2\right)`

Now we want to compute c!


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


`\left(√(3)\right)^2 \ \textgreater \ √(a)=a^{(1)/(2)} \ \textgreater \ \left(3^{(1)/(2)}\right)^2 \ \textgreater \ \mathrm{Apply\:exponent\:rule}: \left(a^b\right)^c=a^(bc) \ \textgreater \ 3^{(1)/(2)\cdot \:2}`


`(1)/(2)\cdot \:2 \ \textgreater \ \mathrm{Multiply\:fractions}: \:a\cdot (b)/(c)=(a\:\cdot \:b)/(c) \ \textgreater \ (1\cdot \:2)/(2) \ \textgreater \ \mathrm{Apply\:rule}\:1\cdot \:a=a`

`(2)/(2) \ \textgreater \ \mathrm{Cancel\:the\:common\:factor:}\:2 \ \textgreater \ 1 \ \textgreater \ 3 \ \textgreater \ √(3-1) \ \textgreater \ √(2)`

Therefore the foci is
`\left(-2+√(2),\:2\right),\:\left(-2-√(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:(1)/(3)\left(x^2+4x\right)+\left(y^2-4y\right)=-(13)/(3)`


`\mathrm{Convert}\:x\:\mathrm{to\:square\:form}`

`(1)/(3)\left(x^2+4x+4\right)+\left(y^2-4y\right)=-(13)/(3)+(1)/(3)\left(4\right)`


`\mathrm{Convert\:to\:square\:form}`

`(1)/(3)\left(x+2\right)^2+\left(y^2-4y\right)=-(13)/(3)+(1)/(3)\left(4\right)`


`\mathrm{Convert}\:y\:\mathrm{to\:square\:form}`

`(1)/(3)\left(x+2\right)^2+\left(y^2-4y+4\right)=-(13)/(3)+(1)/(3)\left(4\right)+4`


`\mathrm{Convert\:to\:square\:form}`

`(1)/(3)\left(x+2\right)^2+\left(y-2\right)^2=-(13)/(3)+(1)/(3)\left(4\right)+4`


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

`(\left(x+2\right)^2)/(3)+(\left(y-2\right)^2)/(1)=1`

Rewrite in the standard form again

`image`


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

Which means the vertices are
`\left(-2+√(3),\:2\right),\:\left(-2-√(3),\:2\right)`

Hope this helps!