【数学I】例題3.3.15：放物線と直線の共有点の個数（One More）★★

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% 例題I3.3.15：放物線と直線の共有点の個数（One More）★★
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次の放物線と直線の共有点の個数を調べよ．ただし，$k$を定数とする．

(1) 放物線$y=3x^2+6x+4$，直線$y=-3x+k$

(2) 放物線$y=x^2-2kx+k^2+5$，直線$y=2x+2$

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% 例題I3.3.15の解答（One More）★★
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(1) $\{\begin{array}{l}y=3x^2+6x+4\cdots(\mathrm{i})\\y=-3x+k\cdots(\mathrm{ii})\end{array}$とする．

(i) と(ii)から$y$を消去すると，$3x^2+6x+4=-3x+k$

整理すると，$3x^2+9x+4-k=0$

この2次方程式の判別式を$D$とすると，

\[
D=9^2-4\cdot 3\cdot(4-k)=81-12(4-k)=12k+33
\]

よって，共有点の個数は，$k$の値によって次のようになる．

\begin{align*}
&12k+33>0\text{のとき，すなわち，}k>-\frac{11}{4}\text{のとき，}&&\text{2個}\\
&12k+33=0\text{のとき，すなわち，}k=-\frac{11}{4}\text{のとき，}&&\text{1個}\\
&12k+33<0\text{のとき，すなわち，}k<-\frac{11}{4}\text{のとき，}&&\text{0個}
\end{align*}

(2) $\{\begin{array}{l}y=x^2-2kx+k^2+5\cdots(\mathrm{i})\\y=2x+2\cdots(\mathrm{ii})\end{array}$とする．

(i) と(ii)から$y$を消去すると，$x^2-2kx+k^2+5=2x+2$

整理すると，$x^2-2(k+1)x+(k^2+3)=0$

この2次方程式の判別式を$D$とすると，

\[
\frac{D}{4}=\{-(k+1)\}^2-1\cdot(k^2+3)=2k-2
\]

よって，共有点の個数は，$k$の値によって次のようになる．

\begin{align*}
&2k-2>0\text{のとき，すなわち，}k>1\text{のとき，}&&\text{2個}\\
&2k-2=0\text{のとき，すなわち，}k=1\text{のとき，}&&\text{1個}\\
&2k-2<0\text{のとき，すなわち，}k<1\text{のとき，}&&\text{0個}
\end{align*}

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% 問題I3.3.15：放物線と直線の共有点の個数（One More）★★
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次の放物線と直線の共有点の個数を調べよ．ただし，$k$を定数とする．

(1) 放物線$y=2x^2+4x-3$，直線$y=-2x+k$

(2) 放物線$y=-x^2+3x+k+1$，直線$y=-x+2$

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% 問題I3.3.15の解答（One More）★★
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(1) $\{\begin{array}{l}y=2x^2+4x-3\cdots(\mathrm{i})\\y=-2x+k\cdots(\mathrm{ii})\end{array}$とする．

(i) と(ii)から$y$を消去すると，$2x^2+4x-3=-2x+k$

整理すると，$2x^2+6x-3-k=0$

この2次方程式の判別式を$D$とすると，

\[
\frac{D}{4}=3^2-2\cdot(-3-k)=9+2(3+k)=2k+15
\]

よって，共有点の個数は，$k$の値によって次のようになる．

\begin{align*}
&2k+15>0\text{のとき，すなわち，}k>-\frac{15}{2}\text{のとき，}&&\text{2個}\\
&2k+15=0\text{のとき，すなわち，}k=-\frac{15}{2}\text{のとき，}&&\text{1個}\\
&2k+15<0\text{のとき，すなわち，}k<-\frac{15}{2}\text{のとき，}&&\text{0個}
\end{align*}

(2) $\{\begin{array}{l}y=-x^2+3x+k+1\cdots(\mathrm{i})\\y=-x+2\cdots(\mathrm{ii})\end{array}$とする．

(i) と(ii)から$y$を消去すると，$-x^2+3x+k+1=-x+2$

整理すると，$x^2-4x-k+1=0$

この2次方程式の判別式を$D$とすると，

\[
\frac{D}{4}=(-2)^2-1\cdot(-k+1)=4+(k-1)=k+3
\]

よって，共有点の個数は，$k$の値によって次のようになる．

\begin{align*}
&k+3>0\text{のとき，すなわち，}k>-3\text{のとき，}&&\text{2個}\\
&k+3=0\text{のとき，すなわち，}k=-3\text{のとき，}&&\text{1個}\\
&k+3<0\text{のとき，すなわち，}k<-3\text{のとき，}&&\text{0個}
\end{align*}

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% 例題I3.3.15の側注：放物線と直線の共有点の個数（One More）★★
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% 側注1
$k$の値によって，直線(ii)が変化する．

% 側注2
放物線と直線が接する$D=0$のときであり，共有点を1個もつ（重解）．

% 側注3
$k$の値によって，放物線(i)が変化する．

% 側注4
放物線と直線が接する．

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% 問題I3.3.15の側注：放物線と直線の共有点の個数（One More）★★
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% 側注1
$k$の値によって，{直線(ii)}が変化する．

% 側注2
放物線と直線が接する．

% 側注3
$k$の値によって，{放物線(i)}が変化する．

% 側注4
放物線と直線が接する．

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% 例題I3.3.15の考え方・補足など：放物線と直線の共有点の個数（One More）★★
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% 考え方
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$y$を消去して導いた2次方程式の判別式を考えることで，共有点の個数を調べる．