Question

How do you find the coordinates of the vertices, foci and the equation of the asymptotes for the hyperbola \[{y^2} = 36 + 4{x^2}\]?

Answer 1

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.