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How to find the canonical equation of the hyperbola if its asymptote has equation $3x + 5y = 0$ and the distance between the foci is equal to 7?

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    Try working backwards: given the equation of a hyperbola, what are the equations of its asymptotes? You should be able to find a relationship between the sets of coefficients between the two equations. – 2017-01-25
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    $y=\pm \frac{b}{a} x$ - ? – 2017-01-25
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    $y=-\frac{3}{5}x$ – 2017-01-25
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    $\frac{x^2}{5^2}-\frac{y^2}{3^2}=1$? But distance between the foci is equal $2\sqrt{34}$, not $7$ :( – 2017-01-25

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Assuming that the hyperbola is meant to be in standard position (otherwise there’s not enough information for a unique solution), you can proceed as follows:

The slopes of the asymptotes of the hyperbola $x^2/a^2-y^2/b=1$ are $\pm b/a$, and the half-focal length $c$ satisfies $c^2=a^2+b^2$. You’ve been given the distance $2c$ and can extract the slope of the asymptote from its equation, so all that’s left is to solve this system of equations for $a$ and $b$. This will basically be a matter of scaling the numerator and denominator of the asymptote’s slope so that the sum of their squares gives the correct focal distance.

In this problem, you’ve already determined that $b/a=-3/5$, so now you need to find a scale factor $\lambda$ such that $(5\lambda)^2+(3\lambda)^2=(7/2)^2$.

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    $(5\lambda)^2+(3\lambda)^2=(7/2)^2$, =>, $\lambda=\pm \frac{7}{2 \cdot \sqrt{34}}$ - ? – 2017-01-26
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    It's amazing, thank you very much! – 2017-01-26