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  1. #1
    Junior Member ihmth's Avatar
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    Equations

    Let S denote the set of points of intersection of the hyperbola xy = 2 and the graph of y = cube root of $\displaystyle x^3 - 20$. How many lines of slope 1 pass through at least one point of S?
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    Quote Originally Posted by ihmth View Post
    Let S denote the set of points of intersection of the hyperbola xy = 2 and the graph of y = cube root of $\displaystyle x^3 - 20$. 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 $\displaystyle y = \sqrt[3]{x^3 - 20}$ ....? 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 $\displaystyle x \, \sqrt[3]{x^3 - 20} = 2 \Rightarrow x^3 (x^3 - 20) = 8 \Rightarrow x^6 - 20 x^3 - 8 = 0$.

    Let $\displaystyle x^3 = w$, say:

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

    Therefore $\displaystyle x^3 = 10 \pm 6 \sqrt{3}$ and so there are two solutions for x. Therefore there are two points in S.
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  3. #3
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    Quote Originally Posted by ihmth View Post
    Let S denote the set of points of intersection of the hyperbola xy = 2 and the graph of y = cube root of $\displaystyle x^3 - 20$. How many lines of slope 1 pass through at least one point of S?
    First calculate the points of intersection:

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

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

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

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

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

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

    To answer your question: There is only one line passing through the whole set of intersection points.
    Attached Thumbnails Attached Thumbnails Equations-ihmth_intersect.jpg  
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    Hello, ihmth

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

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

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

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

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


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


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

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

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    Quote Originally Posted by mr fantastic View Post
    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 $\displaystyle y = \sqrt[3]{x^3 - 20}$ ....? 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 .....
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