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Given that a hyperbola has asymptote $y=0$, passes through the point $(1,1)$ and has axis $y=2x+2$, determine its equation.

The answer arrived at is $\displaystyle{4xy+3y^2+4y-11=0}$. However, I have had no success in reaching it.

I first tried to relate $a$ and $b$ using the point $(1,1)$ to get $$ \frac 1{a^2} - \frac{1}{b^2} = 1 = \frac{a^2b^2}{a^2b^2}$$

Then I changed the subject of the formula $$ x=\frac{a\left(4a+b\sqrt{b^2+4-4a^2}\right)}{b^2-4a^2},\:x=\frac{a\left(4a-b\sqrt{b^2+4-4a^2}\right)}{b^2-4a^2};\quad \:b^2-4a^2\ne \:0 $$ But I didn't find any helful use of that information. Next, I located the centre $(-1,0)$, which is the intersection of the axis and the asymptote. I then related the vertices $$\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = (-1,0)$$ but that gives two equations in four unknowns. I tried to substitute these equations in the canonical equation of a hyperbola but always ended up with a more complex equation that with more than one variable. At this point, I'm out of ideas. How can I approach this problem?

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    $\begingroup$ The axis halves the angle between the asymptotes. So, together with the center $(-1,0)$ you can find it to be $3y+4x+4=0$. Now $y(3y+4x+4)=k$ and using the point $(1,1)$ we find $k=11$. $\endgroup$ Apr 6, 2019 at 7:53
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    $\begingroup$ @Jan-MagnusØkland Thanks, that's the property that I didn't consider. I now know how to proceed. $\endgroup$
    – E.Nole
    Apr 6, 2019 at 7:59

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As pointed out in the comment by Jan-MagnusØkland, solving the problem requires using the property that the axis bisects the angle between the two asymptotes of the hyperbola. We are given one asymptote $y=0$ thus we can find the second one by reflecting $y=0$ across the axis $y=2x+2$. Let's rewrite it in the standard form $2x-y+2=0$.

Let's pick the point $(0,0)$ to reflect. The distance between the axis and the point $(0,0)$ is $$d=\left| \frac{Ax_0 + By_0 +C}{\sqrt{A^2 + B^2}} \right| = \left| \frac{2x_0 -y_0 +2}{\sqrt{(2)^2 + (-1)^2}} \right| = \left| \frac{2\cdot0 -1\cdot0 +2}{\sqrt{5}} \right| = \left| \frac{2}{\sqrt{5}} \right|$$

Therefore, the distance from the symmetric point $(x_0, y_0)$ is also $\frac{2}{\sqrt{5}}$. This gives the equation:

$$ \left| \frac{2x_0 -y_0 +2}{\sqrt{5}} \right| = \frac{2}{\sqrt{5}} \implies \left| {2x_0 -y_0 +2} \right| = 2 \qquad \qquad \qquad (*)$$

Since the point symmetric to $(0,0)$ with respect to the axis $2x-y+2=0$ lies on the perpendicular to $2x-y+2=0$, we can find the gradient of the perpendicular $$ m_1m_2=-1 \implies m_2=\frac{-1}2$$

Thus, together with $(*)$ we have our second (or third) equation $$\frac{y_2-y_1}{x_2-x_1} = \frac{y_0 - 0}{x_0 - 0} = \frac{-1}2 \implies x_0 = -2y_0 \qquad \qquad \qquad (**)$$.

The equation in $(*)$ gives us two cases:

$$\left| {2x_0 -y_0 +2} \right| = 2 \implies \begin{cases} 2x_0 -y_0 +2 = 2,\\ 2x_0 -y_0 +2 = -2 \end{cases}$$ Solving the fist case of $(*)$ simultaneously with $(**)$, we get $(x_0,y_0)$ = $(0,0)$ which is our initial point. So the second case gives us our symmetric point:

$$\begin{cases} 2x_0 -y_0 +2 = -2\\ x_0 = -2y_0 \end{cases} \implies (x_0,y_0) = \left(\frac{-8}{5}, \frac45 \right)$$

That gives us the reflection of the point $(0,0)$ across the line $2x-y+2=0$.Next, we know that the asymptote intersects the axis at the centre of the hyperbola which gives us $c=(-1,0)$. Using $c$ and the symmetric point we just calculated, we find that the equation of the second asymptote is $$y=\frac{-4}{3}x - \frac43 \equiv 3y+4x+4=0 $$

Using the property that the equation of a hyperbola can be given by its asymptotes $$(Ax+By+C)(A_1x+B_1y+C_1)=k$$ We have $$y(3y+4x+4)=k$$ Since the point $(1,1)$ lies on the hyperbola, we get that $k=11$ giving the final answer to be $$4xy+3y^2+4y-11=0$$

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