Why is it okay to take out the constant when finding asymptotes or hyperbola? $\require{cancel}$I have been tasked to find the equation(s) of the asymptotes for the graph of the following hyperbolic equation:$$16x^2-25y^2-400=0$$My teacher explains that you can just take out the constant term to find the asymptotes. For this specific case:$$16x^2-25y^2\cancel{-400}=0\implies16x^2-25y^2=0$$I have graphed both $16x^2-25y^2-400=0$ and $16x^2-25y^2=0$ and confirmed that $16x^2-25y^2=0$ indeed does graph the asymptotes of $16x^2-25y^2-400=0$.

Can someone provide a simple explanation as to why this is the case?
 A: Here's what's happening
The equation of the hyperbola in standard form is
$$b^2x^2 - a^2y^2 -a^2b^2 = 0$$
Divide by $x^2$
$$b^2 - a^2\frac{y^2}{x^2} - \frac{a^2b^2}{x^2} = 0$$
Now, as a point on the hyperbola approaches infinity, we have the following as the curve is continuous
$$\lim_{x \to \infty} \left(b^2 - a^2\frac{y^2}{x^2} - \frac{a^2b^2}{x^2}\right) = 0$$
The constant term will go to zero as x grows large, therefore we have
$$\lim_{x \to \infty} \left(\frac{y}{x}\right) = \pm\frac{b}{a}$$
Hence we have the pair of lines that are asymptotes to the hyperbola are described the pair of lines equation obtained by removing the constant from the equation. This only works in standard form however, and will have toappropriately have to be translated rotated for the general case
A: Asymptotes are essentially the lines passiing through the center of  the hyperbola. So if $$ax^2+by^2+2hxy+2gx+2fy+c=0, h^2>ab~~~~(1)$$ represents a hyperola.
Then by changing $c$ to $c'$, we demand the same quadratic to represent pairs of straight lines such that $\Delta'$ is zero namely
$$\Delta'=2fgh+abc'+af^2+bg^2+c'h^2=0$$ then
$$ax^2+by^2+2hxy+2gx+2fy+c'=0~~~~(2)$$
will be the combines equation of the asymptotes which can be separated out. The question why to change constant only, can be answered by noting that
if hyperbola, (1) can be written as  $(lx+my+n)(px+qy+r)=\pm s^2 \implies XY=\pm s^2$ represents a hyperbola then putting the constant $s=0$, $(lx+my+n)(px+qy+r)=0$, will represent the asymptotes.
Note that $L_1 L_2=\pm s^2$ is the most general equation of a hyperbola, where $L_1$ and $L_2$ are equations of two non-parallel lines.
A: Recall that when you sketch the graphs of
$$\gamma : 16x^2-25y^2-400=0$$
and of
$$\delta : 16x^2-25y^2 = 0$$
you are representing in the Cartesian plane the points whose coordinates solve the above equations.
So when your teacher says that $\delta$ represents the asypmptotes of $\gamma$ he means that the the solutions to the equations $\gamma$ and $\delta$ (the latter being coordinates of the points on the straight lines $y = \pm \frac45 x$) get closer and closer, the larger are the values of $x$ (or $y$) you consider.
To convince yourself of this, you could try to analyze the situation just in the first quadrant. With this limitation, the above mentioned solutions can be expressed in these terms (can you doublecheck that?)
$$y = \frac45 x$$
and
$$y=\frac45 \sqrt{x^2-25}.$$
Now you can ask yourself: for a given $x$, what is the difference between the $y$-coordinate of the points lying on the hyperbola and on the straight line? What we are trying to determine is therefore
$$\Delta = \frac45 x - \frac45 \sqrt{x^2-25}.$$
If you multiply and divide the expression by the sum of the two terms and use the identity $(A+B)(A-B) = A^2-B^2$ (fill in the details yourself) you get
$$\Delta = \frac{20}{x^2+\sqrt{x^2-25}},$$
from which you can see that you can make this difference arbitrarily small, by just taking larger and larger values of $x$.
You could try to generalize these results, and see that the conclusions are not affected by a different choice of the original equation's coefficients.
Hope this helped you.
