Question 1168977
Solution:
The equation of a hyperbola with a conjugate axis along the y-axis has the standard form:
$\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$
where $(h, k)$ is the center of the hyperbola, $2a$ is the length of the transverse axis, and $2b$ is the length of the conjugate axis.

The asymptotes of this hyperbola are given by:
$y - k = \pm \frac{a}{b}(x - h)$

We are given the equations of the asymptotes as $6x - 5y + 30 = 0$ and $6x + 5y - 30 = 0$. We can rewrite these in the slope-intercept form ($y = mx + c$):
$5y = 6x + 30 \implies y = \frac{6}{5}x + 6$
$5y = -6x + 30 \implies y = -\frac{6}{5}x + 6$

The slopes of the asymptotes are $\pm \frac{6}{5}$. Therefore, $\frac{a}{b} = \frac{6}{5}$.

The intersection of the asymptotes gives the center of the hyperbola. We can solve the system of equations:
$y = \frac{6}{5}x + 6$
$y = -\frac{6}{5}x + 6$
Setting them equal: $\frac{6}{5}x + 6 = -\frac{6}{5}x + 6 \implies \frac{12}{5}x = 0 \implies x = 0$.
Substituting $x=0$ into either equation gives $y = 6$.
So, the center of the hyperbola is $(h, k) = (0, 6)$.

We are given that one vertex is at $(0, 7)$. For a hyperbola with a vertical transverse axis and center $(0, 6)$, the vertices are at $(0, k \pm a) = (0, 6 \pm a)$.
Since one vertex is at $(0, 7)$, we have $6 + a = 7$ or $6 - a = 7$.
If $6 + a = 7$, then $a = 1$.
If $6 - a = 7$, then $a = -1$, which is not possible for the length $a$.
Thus, $a = 1$, and $a^2 = 1$.

Now we use the relationship $\frac{a}{b} = \frac{6}{5}$ with $a = 1$:
$\frac{1}{b} = \frac{6}{5} \implies b = \frac{5}{6}$.
So, $b^2 = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$.

The equation of the hyperbola is:
$\frac{(y - 6)^2}{a^2} - \frac{(x - 0)^2}{b^2} = 1$
$\frac{(y - 6)^2}{1} - \frac{x^2}{\frac{25}{36}} = 1$
$(y - 6)^2 - \frac{36x^2}{25} = 1$

Final Answer: The final answer is $\boxed{(y - 6)^2 - \frac{36x^2}{25} = 1}$