Find the equation of the hyperbola whose vertices are

    \[\left( \mathbf{0},\text{ }\pm \mathbf{3} \right)\]

and the eccentricity is

    \[\frac{4}{3}\]

. Also, find the coordinates of its foci.
Find the equation of the hyperbola whose vertices are

    \[\left( \mathbf{0},\text{ }\pm \mathbf{3} \right)\]

and the eccentricity is

    \[\frac{4}{3}\]

. Also, find the coordinates of its foci.

Given: Vertices are

    \[\left( \mathbf{0},\text{ }\pm \mathbf{3} \right)\]

and the eccentricity is

    \[\frac{4}{3}\]

Need to find: The equation of the hyperbola and coordinates of foci.

Let, the equation of the hyperbola be:

    \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\]

Vertices are

    \[\left( \mathbf{0},\text{ }\pm \mathbf{3} \right)\]

, that means, a =

    \[3\]

Given eccentricity

    \[e=\frac{4}{3}\]

We know that,

    \[<span class="ql-right-eqno"> (1) </span><span class="ql-left-eqno">   </span><img src="https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-bd33695f6367bc6206efefb53361451a_l3.png" height="454" width="234" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*} & e=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}} \\ & Therefore, \\ & \sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}=\frac{4}{3} \\ & 1+\frac{{{b}^{2}}}{{{a}^{2}}}=\frac{16}{9} \\ & \frac{{{b}^{2}}}{{{a}^{2}}}=\frac{16}{9}-1=\frac{7}{9} \\ & {{b}^{2}}=\frac{7}{9}{{a}^{2}}=7 \\ \end{align*}" title="Rendered by QuickLaTeX.com"/>\]

So, the equation of hyperbola is,

    \[<span class="ql-right-eqno"> (2) </span><span class="ql-left-eqno">   </span><img src="https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d9fb02f5e74fc8aeb035d134110c642f_l3.png" height="199" width="503" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*}</strong> <strong>  & \frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1 \\ </strong> <strong> & \frac{{{x}^{2}}}{9}-\frac{{{y}^{2}}}{7}=1 \\ </strong> <strong>\end{align*}" title="Rendered by QuickLaTeX.com"/>\]

So, the co-ordinates of the foci are

    \[(\pm ae,0)=(\pm 4,0)\]