# How to find the equation of a hyperbola given the asymptote, equation of axis and a point

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?

• 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 $k=11$. Apr 6, 2019 at 7:53
• @Jan-MagnusØkland Thanks, that's the property that I didn't consider. I now know how to proceed. Apr 6, 2019 at 7:59

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 $$4xy+3y^2+4y-11=0$$