So, you know that the equation is going to look like
$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.$$
You already figured out that $a=-b$. So now you need to figure out what values of $a$ and $b$ will make the foci be at the given point $(4,0)$ and $(-4,0)$.
Since you know that $a=-b$, then you know that $a^2=b^2$, so you can rewrite the equation as
$$\frac{x^2}{a^2} - \frac{y^2}{a^2} = 1$$
or, by clearing denominators (multiplying through by $a^2$)
$$x^2 - y^2 = a^2.$$
Now: where does the hyperbola intersect the $x$ axis? If you have a point $(X,0)$ in the hyperbola, then plugging it in you will have
$$X^2 = a^2.$$
So if you can determine where the hyperbola intersects the $x$-axis, you can use that to figure out the value of $a$, and so the value of $b$, and so the equation for the hyperbola.
Note: There are many, many, many ways to finish the problem, depending on what you know about hyperbolas in general (which is precisely why I asked you in comments what you knew).
Added. One possible way to finish the problem is to use the eccentricity of the hyperbola. The distance from the center to the foci is $a\varepsilon$, where $\varepsilon$ is the eccentricity of the hyperbola. The eccentricity is always equal to
$$\varepsilon = \sqrt{ 1 + \frac{b^2}{a^2}}$$
so here, since $a=b$, that means that the eccentricity is $\sqrt{2}$. Therefore, since the distance from the center to the foci is $4$, this tells you that $4 = a\varepsilon = a\sqrt{2}$. You now know what $a$ is, and thus, you know what $a^2=b^2$ is, and so you know the equation of the hyperbola.
Added 2. Another possibility you discuss in the comments. You have $a$ is the distance from the center to the vertices; $b$ is the distance from the center to the conjugate axis. These distances satisfy $c^2 = a^2+b^2$, where $c$ is the value $4$ that you have. Since $a^2=b^2$, then you have $c^2=2a^2$, or $c = \sqrt{2a^2} = \sqrt{2}|a| = a\sqrt{2}$. You already know what $c$ is, so now you can solve for $a$. Once you know $a$, you also know $b$ (since $a=-b$) which gives you the information you need.
From the two tangents you can find the parabola’s axis direction. It’s the diagonal of the paralellogram formed by the tangents and their intersection point. These two tangents intersect at the origin, so the axis direction is $\mathbf v = (1,1)+(1,0)=(2,1)$.
You can now use the reflective property of the parabola to find its focus. The direction vector $\mathbf v$ is reflected to $(-2,1)$ at the point $(1,0)$ and to $(1,2)$ at the point $(1,1)$, so the parabola’s focus is the intersection of the lines $(1,0)+\lambda(-2,1)$ and $(1,1)+\mu(1,2)$, which can be found via cross products of homogeneous coordinates: $$((1,0,1)\times(-2,1,0))\times((1,1,1)\times(1,2,0))=(-3,-1,-5),$$ or $F=\left(\frac35,\frac15\right)$ in inhomogeneous Cartesian coordinates.
The feet of perpendiculars dropped from the focus to tangent lines lie on the tangent to the parabola’s vertex. The point $\left(\frac35,0\right)$ on this line is easily found by inspection. Knowing also that this line is perpendicular to the parabola’s axis, we can form its point-normal equation: $(2,1)\cdot(x,y)-(2,1)\cdot\left(\frac35,0\right)=0$ or $$2x+y-\frac65=0.$$ This line is halfway between the focus and directrix, so we offset the left-hand side by $(2,1)\cdot F-\frac65 = \frac15$ resulting in the equation $$2x+y=1$$ for the directrix. You now have the necessary information to use the form of the equation of a parabola in your question.
Alternatively, once you have the axis direction, you know that the equation will be of the form $$(x-2y)^2+ax+by+c=0.\tag{*}$$ The normals to this curve can be found by differentiation: $(a+2x-4y,b-4x+8y)$. If the tangent at a point has direction $(\lambda,\mu)$ the condition that the normal is orthogonal to the tangent can be expressed as $$(\lambda,\mu)\cdot(a+2x-4y,b-4x+8y)=0.$$ Plugging in the known points and tangents will give you a system of linear equations in $a$ and $b$. Once you’ve solved for those, you can plug one of the points into equation (*) to find $c$.
Best Answer
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 $$4xy+3y^2+4y-11=0$$