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tamaranim1 [39]
3 years ago
14

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

Mathematics
1 answer:
Xelga [282]3 years ago
8 0
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)

\frac{\left(x-h\right)^2}{a^2}-\frac{\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)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}1
\frac{1}{1}\left(x^2-4x\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}

\mathrm{Convert}\:x\:\mathrm{to\:square\:form}
\frac{1}{1}\left(x^2-4x+4\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)

\mathrm{Convert\:to\:square\:form}
\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)

\mathrm{Convert}\:y\:\mathrm{to\:square\:form}
\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y^2+4y+4\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right)

\mathrm{Convert\:to\:square\:form}
\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y+2\right)^2=-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right)

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

Now rewrite in hyperbola standardform
\frac{\left(x-2\right)^2}{1^2}-\frac{\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
\sqrt{1^2+3^2} \ \textgreater \  \mathrm{Apply\:rule}\:1^a=1 \ \textgreater \  1^2 = 1 \ \textgreater \  \sqrt{1+3^2}

3^2 = 9 \ \textgreater \  \sqrt{1+9} \ \textgreater \  \sqrt{10}

Therefore the hyperbola foci is at \left(2+\sqrt{10},\:-2\right),\:\left(2-\sqrt{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!

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