Question

Find the vertex and foci hyperbola y^2-x^2 = 81

Answer 1

The standard equation of a hyperbola is expressed as

`\begin{gathered} (\left(y-k\right)^2)/(a^2)-(\left(x-h\right)^2)/(b^2)=1\text{ ---- equation 1} \\ \text{where} \\ (h,\text{ k) is the coordinate of its center} \\ a\text{ is the axis} \\ b\text{ is the conjugate axis} \end{gathered}`

Given the equation of the hyperbola to be

`y^2-x^2=81\text{ ---- equation 2}`

Express equation 2 in a similar form as equation 1.

Thus,

`\begin{gathered} y^2-x^2=81 \\ \text{divide both sides of the equation by 81} \\ (y^2-x^2)/(81)=(81)/(81) \\ \Rightarrow(y^2)/(9^2)-(x^2)/(9^2)=1\text{ ---- equation 3} \end{gathered}`

In comparison with equation 1, we can conclude that

`\begin{gathered} a=9 \\ b=9 \end{gathered}`

Vertices of the hyperbola:

The vertices of the hyperbola are expressed as

`\mleft(h,k+a\mright),\: \mleft(h,k-a\mright)`

where

`h=0`

The vertices of the hyperbola are evaluated to be

`\begin{gathered} (h,\: k+a)\Rightarrow(0,\text{ 0+9)=(0, 9)} \\ (h,\: k-a)\Rightarrow(0,0-9)=(0,-9) \end{gathered}`

Hence, the vertices of the hyperbola are

`(0,9),\text{ (0, -9)}`

Foci of the hyperbola:

The foci of the hyperbola are expressed as

`\begin{gathered} \mleft(h,k+c\mright),\: \mleft(h,k-c\mright) \\ \text{where c }\mathrm{\: }is\: the\text{ distance from the center (h,k) to a focus} \\ c\text{ is evaluated as } \\ c=√(a^2+b^2) \end{gathered}`

Evaluating c gives

`\begin{gathered} c=√(a^2+b^2) \\ =\sqrt[]{9^2+9^2} \\ =\sqrt[]{81+81} \\ =\sqrt[]{162} \\ c=9\sqrt[]{2} \end{gathered}`

Thus, the foci are evaluated as

`\begin{gathered} (h,k+c)\Rightarrow(0,\text{ 0+9}\sqrt[]{2})=(0,9\sqrt[]{2}) \\ (h,k-c)\Rightarrow(0,\text{ 0-9}\sqrt[]{2})=(0,-9\sqrt[]{2)} \end{gathered}`

Hence, the foci of the hyperbola are

`\mleft(0,\: 9√(2)\mright),\: \mleft(0,\: -9√(2)\mright)`