How to turn a line and hyperbola intersection into a quadratic equation

Key concept: This is a neat algebra setup: put the line and hyperbola equal to each other, clear the denominator, and the intersection points fall out from the quadratic you build.

Step by step

Let the hyperbola be

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

and the line be

$y=8x-32$

Step 1: Set the equations equal

Since the intersection points have the same $y$-value,

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

Step 2: Clear the denominator

Multiply both sides by $x-4$:

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

You can simplify first:

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

or, using the equivalent form shown in the working,

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

then

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

Step 3: Expand and rearrange

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

$4x^2-32x+63=0$

So the quadratic equation is

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

Its solutions are the $x$-coordinates of the intersection points.

Pitfall alert

A frequent mistake is forgetting to multiply every term by the denominator. If you only clear one side of the equation, the quadratic won’t come out correctly. Another common slip is leaving the equation in a form with fractions, even though the prompt asks for $ax^2+bx+c=0$.

Try different conditions

If the line had a different gradient, you’d still set the two expressions for $y$ equal and multiply through by the denominator of the hyperbola. The final quadratic would change, but the method stays identical: substitute, clear fractions, expand, and collect terms.

Further reading

simultaneous equations, quadratic equation, hyperbola