Question

Find the vertices and foci of the hyperbola. 9x2 − y2 − 36x − 4y + 23 = 0

Answer 1

Hey there, hope I can help!

NOTE: Look at the image/images for 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\:standard\:equation\:for\:a\:right-left\:facing:H}`
with the center of (h, k), semi-axis a and semi-conjugate - axis b.
NOTE: H = hyperbola


`9x^2-y^2-36x-4y+23=0 \ \textgreater \ \mathrm{Subtract\:}23\mathrm{\:from\:both\:sides}`

`9x^2-36x-4y-y^2=-23`


`\mathrm{Factor\:out\:coefficient\:of\:square\:terms}`

`9\left(x^2-4x\right)-\left(y^2+4y\right)=-23`


`\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9`

`\left(x^2-4x\right)-(1)/(9)\left(y^2+4y\right)=-(23)/(9)`


`\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}1`

`(1)/(1)\left(x^2-4x\right)-(1)/(9)\left(y^2+4y\right)=-(23)/(9)`


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

`(1)/(1)\left(x^2-4x+4\right)-(1)/(9)\left(y^2+4y\right)=-(23)/(9)+(1)/(1)\left(4\right)`


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

`(1)/(1)\left(x-2\right)^2-(1)/(9)\left(y^2+4y\right)=-(23)/(9)+(1)/(1)\left(4\right)`


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

`(1)/(1)\left(x-2\right)^2-(1)/(9)\left(y^2+4y+4\right)=-(23)/(9)+(1)/(1)\left(4\right)-(1)/(9)\left(4\right)`


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

`(1)/(1)\left(x-2\right)^2-(1)/(9)\left(y+2\right)^2=-(23)/(9)+(1)/(1)\left(4\right)-(1)/(9)\left(4\right)`


`\mathrm{Refine\:}-(23)/(9)+(1)/(1)\left(4\right)-(1)/(9)\left(4\right) \ \textgreater \ (1)/(1)\left(x-2\right)^2-(1)/(9)\left(y+2\right)^2=1 \ \textgreater \ Refine`

`(\left(x-2\right)^2)/(1)-(\left(y+2\right)^2)/(9)=1`

Now rewrite in hyperbola standardform

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


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

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

Now we must compute c

`√(1^2+3^2) \ \textgreater \ \mathrm{Apply\:rule}\:1^a=1 \ \textgreater \ 1^2 = 1 \ \textgreater \ √(1+3^2)`


`3^2 = 9 \ \textgreater \ √(1+9) \ \textgreater \ √(10)`

Therefore the hyperbola foci is at
`\left(2+√(10),\:-2\right),\:\left(2-√(10),\:-2\right)`

For the vertices we have
`\left(2+1,\:-2\right),\:\left(2-1,\:-2\right)`

Simply refine it

`\left(3,\:-2\right),\:\left(1,\:-2\right)`
Therefore the listed coordinates above are our vertices

Hope this helps!