How to find the equation, asymptotes, and intersection setup for a hyperbola

Key takeaway: This question links three skills: reading a hyperbola from information, finding its asymptotes, and turning an intersection into a quadratic. It rewards clean substitution more than fancy tricks.

(a) Find $p$ and $q$

The hyperbola is

$y=\frac{p}{x+q}$

It passes through $(6,1)$, so

$1=\frac{p}{6+q}$

which gives

$p=6+q$

It also intercepts the vertical axis at $y=-0.5$, so when $x=0$:

$-0.5=\frac{p}{q}$

Now use the values given in the working:

$p=2,\quad q=-4$

Check:

$\frac{2}{6-4}=1$

and

$\frac{2}{-4}=-0.5$

(b) Asymptotes

For

$y=\frac{2}{x-4}$

the vertical asymptote is where the denominator is zero:

$x=4$

The horizontal asymptote is

$y=0$

(c) Form the quadratic for the intersections with $y=8x-32$

Set the two equations equal:

$8x-32=\frac{2}{x-4}$

Clear the denominator by multiplying through by $x-4$:

$(8x-32)(x-4)=2$

A simplified equivalent route is

$4x-16=\frac{1}{x-4}$

so

$(4x-16)(x-4)=1$

Expand:

$4x^2-32x+64=1$

Rearrange:

$\boxed{4x^2-32x+63=0}$

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Pitfalls the pros know 👇 Two things go wrong here a lot. First, students sometimes call the vertical asymptote $y=4$ instead of $x=4$ because they mix up which coordinate comes from the denominator. Second, when forming the quadratic, it’s easy to forget to expand both brackets fully before collecting terms.

What if the problem changes? If the intercept on the vertical axis were different, you’d still solve two equations for $p$ and $q$ by substituting the known point and the $y$-intercept. Once the hyperbola is known, the asymptotes still come straight from the denominator and the intersection step still starts by setting the line equal to the curve.

Tags: asymptote, hyperbola, quadratic equation