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**Hint:**We know the standard equation of hyperbola with a horizontal transverse axis is \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\] and also we know the standard equation of hyperbola with a vertical transverse axis is \[\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1\]. In both the cases the centre is origin. We convert the given equation into one of the form and then we can find the vertices, foci and asymptotes.

**Complete step by step solution:**

Given \[{y^2} = 36 + 4{x^2}\]

Rearranging we have,

\[{y^2} - 4{x^2} = 36\]

Divide the whole equation by 36 we have,

\[\dfrac{{{y^2}}}{{36}} - \dfrac{{4{x^2}}}{{36}} = \dfrac{{36}}{{36}}\]

\[\dfrac{{{y^2}}}{{36}} - \dfrac{{{x^2}}}{9} = 1\]

Since 36 and 9 is a perfect square, we can rewrite it as

\[\dfrac{{{y^2}}}{{{6^2}}} - \dfrac{{{x^2}}}{{{3^2}}} = 1\]

It is of the form \[\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1\].

Comparing the obtained solution with general solution we have \[a = 6\] and \[b = 3\]

Thus we standard equation of hyperbola with a vertical transverse axis and with centre at origin.

The hyperbola is centred at the origin so the vertices serve as the y-intercept of the graph. Now to find the vertices set \[x = 0\] and solve for ‘y’.

\[\dfrac{{{y^2}}}{{{6^2}}} - \dfrac{{{0^2}}}{{{3^2}}} = 1\]

\[\dfrac{{{y^2}}}{{{6^2}}} = 1\]

\[{y^2} = {6^2}\]

Taking square root on both sides we have,

\[y = \pm 6\]

Thus the vertices are \[(0,6)\] and \[(0, - 6)\].

The foci are located at \[(0, \pm c)\].

We know that \[c = \sqrt {{a^2} + {b^2}} \]

\[\begin{gathered}

c = \pm \sqrt {{6^2} + {3^2}} \\

c = \pm \sqrt {36 + 9} \\

c = \pm \sqrt {45} \\

c = \pm \sqrt {9 \times 5} \\

c = \pm 3\sqrt 5 \\

\end{gathered} \]

Hence the foci are \[(0, \pm 3\sqrt 5 )\].

Now the equation of asymptotes will be \[y = \pm \left( {\dfrac{a}{b}} \right)x\]

\[\begin{gathered}

y = \pm \left( {\dfrac{6}{3}} \right)x \\

y = \pm 2x \\

y \mp 2x = 0 \\

\end{gathered} \]

**Hence the equation of asymptotes are \[y + 2x = 0\] and \[y - 2x = 0\].**

**Note:**If they ask us to find eccentricity for the same problem, we know the formula for eccentricity of a hyperbola is \[e = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} \]. Substituting the value of ‘a’ and ‘b’ we will get the answer.

\[\begin{gathered}

e = \sqrt {1 + \dfrac{{{3^2}}}{{{6^2}}}} \\

e = \sqrt {1 + \dfrac{9}{{36}}} \\

e = \sqrt {\dfrac{{36 + 9}}{{36}}} \\

e = \sqrt {\dfrac{{45}}{{36}}} \\

e = \dfrac{{3\sqrt 5 }}{6} \\

\end{gathered} \]

Thus the eccentricity is \[\dfrac{{3\sqrt 5 }}{6}\].

Careful in the substitution and calculation part.

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