\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)

2012.3.3 Question 3

1. Let the two curves be \(\Gamma _1: y = 4 - x^2\) and \(\Gamma _2: x = - \frac {y^2}{m} + \frac {k}{m}\).

   For the first curve, its \(y\)-intercept is \(4\), and its \(x\)-intercept is \(\pm 2\).

   For the second curve, its \(y\)-intercept is \(\pm \sqrt {k}\) (if \(k \geq 0\)), and its \(x\)-intercept is \(\frac {k}{m}\).

   1. Since \(k < 0\), we must have \(\frac {k}{m} < 0\) as well, and hence the curves must look as follows:

      

   2. Since \(0 < k < 16\), \(\Gamma _2\) must have a \(y\)-intercept less than that of \(\Gamma _1\). Since \(\frac {k}{m} < 2\), \(\Gamma _2\) must have the \(x\)-intercept to the left of \((2, 0)\). Hence, the curves must look as follows:

      

   3. Since \(k > 16\), \(\Gamma _2\) must have a \(y\)-intercept greater than that of \(\Gamma _1\). Since \(\frac {k}{m} > 2\), \(\Gamma _2\) must have the \(x\)-intercept to the right of \((2, 0)\). Hence, the curves must look as follows:

      

   4. Since \(k > 16\), \(\Gamma _2\) must have a \(y\)-intercept greater than that of \(\Gamma _1\). Since \(\frac {k}{m} < 2\), \(\Gamma _2\) must have the \(x\)-intercept to the left of \((2, 0)\). Hence, the curves must look as follows:

      

2. Since \(y = y\), we must have \[ 12x = k - (4 - x^2)^2 = k - 16 + 8 x^2 - x^4, \] and hence \[ x^4 - 8x^2 + 12 x + 16 - k = 0, \] as desired.

   For the first curve, we have \[ \DiffFrac {y}{x} = -2x, \] and applying implicit differentiation on both sides of the second equation, we must have \[ 12 = - 2 y \DiffFrac {y}{x}, \] and hence \[ 12 = (-2y) \cdot (-2x), \] which gives \(xy = 3\) for the point where the curves touch.

   Hence, \[ \frac {3}{a} = 4 - a^2, \] and this gives \[ a^3 - 4a + 3 = 0 \] as desired.

   Notice that \[ a^3 - 4a + 3 = (a - 1)(a^2 + a - 3), \] and hence the three solutions to \(a\) are \[ a_1 = 1, a_{2, 3} = \frac {-1 \pm \sqrt {1 + 12}}{2} = \frac {-1 \pm \sqrt {13}}{2}. \]

   From the first equation, we must have \begin {align*} k & = a^4 - 8a^2 + 12a + 16 \\ & = a (a^3 - 4a + 3) - 4a^2 + 9a + 16 \\ & = a \cdot 0 - 4a^2 + 9a + 16 \\ & = -4a^2 + 9a + 16, \end {align*}

   as desired.

   For \(a = 1\), \(k = -4 \cdot 1^2 + 9 \cdot 1 + 16 = -4 + 9 + 16 = 21\), and \(\frac {k}{m} = \frac {21}{12} < 2\), so (d) arises.

   When \(a_{2, 3} = \frac {-1 \pm \sqrt {13}}{2}\), we have \(a^2 + a - 3 = 0\), and hence \[ k = -4a^2 + 9a + 16 = -4(a^2 + a - 3) + 13a + 4 = 13a + 4. \]

   For \(a_2 = \frac {-1 + \sqrt {13}}{2}\), we have \[ k = \frac {-13 + 13\sqrt {13}}{2} + 4 = \frac {-5 + 13\sqrt {13}}{2}. \]

   Since \(13\sqrt {13} > 13 \cdot 3 = 39\), we must have \(-5 + 13\sqrt {13} > 34\), and hence \(k > \frac {34}{2} = 17 > 16\).

   We also have \(13 \sqrt {13} < 13 \cdot 4 = 52\), and hence \(-5 + 13\sqrt {13} < 47\), and hence \(k < \frac {47}{2}\), which means \[ \frac {k}{m} < \frac {47}{2 \cdot 12} = \frac {47}{24} < 2, \] so case (d) arises.

   For \(a_3 = \frac {-1 - \sqrt {13}}{2}\), we have \(k = \frac {-13 - 13\sqrt {13}}{2} + 4 = \frac {-5 - 13\sqrt {13}}{4} < 0\), and so (a) arises.