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Question 1168977: Conjugate axis along the y-axis, ome vertex at (0,7), asymptotes are 6x-5y+30=0 and 6x+5y-30=0. Need help finding the Equation of this hyperbola. How?
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! 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 $\boxed{(y - 6)^2 - \frac{36x^2}{25} = 1}$
Answer by ikleyn(52879)  (Show Source):
You can put this solution on YOUR website! .
Conjugate axis along the y-axis, one vertex at (0,7), asymptotes are 6x-5y+30=0 and 6x+5y-30=0.
Need help finding the Equation of this hyperbola. How?
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Regarding this post, I have two notices.
First notice is that the problem's formulation is INCORRECT and SELF-CONTRADICTORY.
It describes a situation which NEVER may happen in reality.
Indeed, the problem says that conjugate axis of the hyperbola is along y-axis,
one vertex is (0,7) and the asymptotes are 6x-5y+30 = 0 and 6x+5y-30 = 0.
From equations, the center of the hyperbola is (x,y) = (0,6).
Together with the information about " one vertex ", it means that transverse axis is vertical y-axis,
and it contradicts that the conjugate axis is y-axis, as stated in the post.
It disproves the problem, kills it to the death and ruins it into dust.
Second notice is that the solution in the post by @CPhill is TOTALLY and FATALLY incorrect,
starting from his first two lines, where he writes an equation of the hyperbola in WRONG FORM.
So, the problem itself is  , as well as its " solution " produced and presented by @CPhill.
For the peace in your mind, my dear reader, ignore both the problem and its " solution " by @CPhill.
A right place for such compositions is a trash bin.
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