1. ## Equations

Let S denote the set of points of intersection of the hyperbola xy = 2 and the graph of y = cube root of $x^3 - 20$. How many lines of slope 1 pass through at least one point of S?

2. Originally Posted by ihmth
Let S denote the set of points of intersection of the hyperbola xy = 2 and the graph of y = cube root of $x^3 - 20$. How many lines of slope 1 pass through at least one point of S?
I'm not sure I understand the last bit "How many lines of slope 1 pass through at least one point of S?" .....

You can draw a line of any desired gradient (including m = 1) through any given point (including points in S).

So I would've thought the answer would be equal to the number of points in S .....

Do you mean how many tangents to either xy = 2 or $y = \sqrt[3]{x^3 - 20}$ ....? Or, perhaps how many lines passing through two of the points in S ....?

Anyway, you certainly need to know how many points are in S ......

Solve $x \, \sqrt[3]{x^3 - 20} = 2 \Rightarrow x^3 (x^3 - 20) = 8 \Rightarrow x^6 - 20 x^3 - 8 = 0$.

Let $x^3 = w$, say:

$w^2 - 20 w - 8 = 0 \Rightarrow w = 10 \pm \sqrt{108} = 10 \pm 6 \sqrt{3}$.

Therefore $x^3 = 10 \pm 6 \sqrt{3}$ and so there are two solutions for x. Therefore there are two points in S.

3. Originally Posted by ihmth
Let S denote the set of points of intersection of the hyperbola xy = 2 and the graph of y = cube root of $x^3 - 20$. How many lines of slope 1 pass through at least one point of S?
First calculate the points of intersection:

$\frac2x = \sqrt[3]{x^3-20}$. Cube both sides, multiply by x³. You'll get an equation:

$x^6-20x^3-8 = 0$ . Use the substitution t = x³

$t^2-20t-8=0~\implies~t=10\pm 6\sqrt{3}$ which will give only 2 x-values. Plug in these x-values into the equation of the hyperbola:

$P_1\left(1+\sqrt{3} , \sqrt{3}-1\right), P_2\left(1-\sqrt{3} , -\sqrt{3}-1\right)$

Now use point-slope-formula to calculate the equations of the lines passing through $P_1$ or through $P_2$.

You'll get in both cases: $y = x-2$

To answer your question: There is only one line passing through the whole set of intersection points.

4. Hello, ihmth

Let $S$ denote the set of points of intersection of the hyperbola $xy = 2$
. . and the graph of $y \:=\:\sqrt[3]{x^3 - 20}$

How many lines of slope 1 pass through at least one point of $S$ ?
We have: . $\begin{array}{ccccccc}xy \; =\;2 & \quad\Rightarrow\quad & y \; = \;\frac{2}{x} & {\color{blue}[1]}\\
y \; = \; \sqrt[3]{x^3-20} & \quad\Rightarrow\quad & x^3-y^3 \; = \; 20 & {\color{blue}[2]}\end{array}$

Substitute [1] into [2]: . $x^3 - \left(\frac{2}{x}\right)^3 \:=\:20\quad\Rightarrow\quad x^6 - 20x^3 - 8 \;=\;0$

Quadratic Formula: . $x^3 \;=\;\frac{20 \pm\sqrt{432}}{2} \;=\;10 \pm 6\sqrt{3} \;=\;(1\pm\sqrt{3})^3$

. . Hence: . $x \:=\:1 \pm\sqrt{3}\quad\Rightarrow\quad y \:=\:-1\pm\sqrt{3}$

Therefore: . $S$ has two points: . $P\left(1\!+\!\sqrt{3},\:\text{-}1\!+\!\sqrt{3}\right),\;Q\left(1\!-\!\sqrt{3},\:\text{-}1\!-\!\sqrt{3}\right)$

Since the slope of $PQ$ is: . $m_{_{PQ}} \;=\;\frac{(\text{-}1-\sqrt{3}) - (\text{-}1+\sqrt{3})}{(1-\sqrt{3}) - (1+\sqrt{3})} \;=\; \frac{-2\sqrt{3}}{-2\sqrt{3}} \;=\;1$

. . there is one line with slope 1 that passes through $P\text{ or }Q.$

5. Originally Posted by mr fantastic
I'm not sure I understand the last bit "How many lines of slope 1 pass through at least one point of S?" .....

You can draw a line of any desired gradient (including m = 1) through any given point (including points in S).

So I would've thought the answer would be equal to the number of points in S .....

Do you mean how many tangents to either xy = 2 or $y = \sqrt[3]{x^3 - 20}$ ....? Or, perhaps how many lines passing through two of the points in S ....?

[snip]
I see nothing to change my opinion here ..... The question is open to several interpretations .....