反殷芳

由於眾所周知的原因，維基百科被查了水錶，所以暫居這裡。

一些反正也記不住不過可以自行推導的公式

$A\sin x+A\cos x+B\sin x\cos x={\frac {B}{2}}t^{2}+At-{\frac {B}{2}}$，其中$t=\sin x+\cos x={\sqrt {2}}\sin \left(x+{\frac {\pi }{4}}\right)$

${\begin{aligned}A\sin ^{2}x+B\cos ^{2}x+C\sin x\cos x&={\frac {C}{2}}\sin 2x+{\frac {B-A}{2}}\cos 2x+{\frac {B+A}{2}}\\&={\frac {\sqrt {A^{2}+B^{2}+C^{2}-2AB}}{2}}\sin \left(2x+\arctan {\frac {B-A}{C}}\right)\\\end{aligned}}$

對戴芊問題的回答：函數f(x)=sin(kx)cos(jx)的周期

2018年1月17日手稿（已與戴）

（原手稿有不嚴謹處）

顯然$f(x)=\sin kx\cos jx={\frac {1}{2}}\left[\sin(k+j)x+\sin(k-j)x\right]$

令$p=k+j$，$q=k-j$，則有$f(x)={\frac {1}{2}}\left(\sin px+\sin qx\right)$

$\Rightarrow f\left(x+{\frac {2\pi }{n}}\right)={\frac {1}{2}}\left[\sin \left(px+{\frac {2p\pi }{n}}\right)+\sin \left(qx+{\frac {2q\pi }{n}}\right)\right]$

∵sin(x)的周期為$2\pi$，∴${\frac {2p\pi }{n}}$與${\frac {2q\pi }{n}}$皆為$2\pi$的整數倍，即${\frac {p}{n}},{\frac {q}{n}}\in \mathbb {Z}$

情況一：若$p,q\in \mathbb {Z}$，則$n\mid p,\ n\mid q$. 又由於T為最小正周期，$n=\gcd(\left\vert p\right\vert ,\left\vert q\right\vert )$。故$T={\frac {2\pi }{\gcd(\left\vert k+j\right\vert ,\left\vert k-j\right\vert )}}$

情況二：若$\left(p\notin \mathbb {Z} \right)\lor \left(q\notin \mathbb {Z} \right)$，

1) 若$\left[\left(p\in \mathbb {R} \setminus \mathbb {Q} \right)\lor \left(q\in \mathbb {R} \setminus \mathbb {Q} \right)\right]\land \left(\nexists n\in \mathbb {Q} ,\ n={\frac {p}{q}}\right)\Rightarrow f(x)$為非周期函數

2) 若$\left(p\in \mathbb {Q} \right)\land \left(q\in \mathbb {Q} \right)$，不妨令$p={\frac {a_{1}}{b_{1}}}$，$q={\frac {a_{2}}{b_{2}}}$，$a_{1},a_{2},b_{1},b_{2}\in \mathbb {Z} ,\ \gcd(a_{1},b_{1})=\gcd(a_{2},b_{2})=1$

則${\frac {a_{1}}{b_{1}n}},\ {\frac {a_{2}}{b_{2}n}}\in \mathbb {Z} \Rightarrow {\frac {1}{a_{1}n}},\ {\frac {1}{b_{1}n}}\in \mathbb {Z}$

易知$n={\frac {1}{\operatorname {lcm} (\left\vert b_{1}\right\vert ,\left\vert b_{2}\right\vert )}}\Rightarrow T=2\operatorname {lcm} (\left\vert b_{1}\right\vert ,\left\vert b_{2}\right\vert )\pi$

3) 若$\left(p\in \mathbb {R} \setminus \mathbb {Q} \right)\land \left(q\in \mathbb {R} \setminus \mathbb {Q} \right)\land \left(\exists n\in \mathbb {Q} ,\ n={\frac {p}{q}}\right)\Rightarrow f(x)$為周期函數。將$x$換元，即可如情況2處理。

對戴芊問題的回答：數列通項的不動點求法

首先要明白這樣一個出題的邏輯：命題者肯定是先列出一個不那麼複雜的數列關係（譬如某多項式倒數呈等差數列），然後將其進行處理，使其變得複雜。因此，我們宜從結果開始，倒推出這一方法適用的題型。

$\left\{{\frac {1}{a_{n}-x}}\right\}$為等差數列……①

$\Leftarrow {\frac {1}{a_{n+1}-x}}-{\frac {1}{a_{n}-x}}=d$……②

$\Leftarrow a_{n+1}={\frac {(xd+1)a_{n}-x^{2}d}{da_{n}-xd+1}}$……③

令$m_{1}=xd+1$，$n_{1}=-x^{2}d$，$m_{2}=d$，$n_{2}=-xd+1$。則顯然①適用於形如$a_{n+1}={\frac {m_{1}a_{n}+n_{1}}{m_{2}a_{n}+n_{2}}}$的遞推關係。由於將x同時帶入③中的$a_{n+1},\ a_{n}$，左右相等恆成立，也就證明了不動點法求解的正確性。

一

一

$\lim _{x\to 0}(x+e^{x})^{\frac {1}{x}}=e^{2}$

解：令$y={\frac {1}{x}}\ln(x+e^{x})$，注意到$x\rightarrow 0$時，$x+e^{x}\rightarrow 1$，故$\ln(x+e^{x})\sim x+e^{x}-1$.

$\lim _{x\to 0}y=\lim _{x\to 0}{\frac {x+e^{x}-1}{x}}=\lim _{x\to 0}1+e^{x}=2$，即$\lim _{x\to 0}(x+e^{x})^{\frac {1}{x}}=e^{2}$

二

$\lim _{x\to 0^{+}}\sin \arctan {\frac {1}{x}}=1$

解：令$t={\frac {1}{x}}$，則$x\rightarrow 0^{+}$時，$t\rightarrow +\infty$

故$\lim _{x\to 0^{+}}\sin \arctan {\frac {1}{x}}=\sin \left(\lim _{x\to 0^{+}}\arctan {\frac {1}{x}}\right)=\sin \left(\lim _{t\to +\infty }\arctan t\right)=\sin {\frac {\pi }{2}}=1$.

二

一

$\int _{0}^{\infty }{\frac {\mathrm {d} x}{\left(1+x\right)^{3}}}={\frac {2\pi }{3{\sqrt {3}}}}$

解: 不妨設${\frac {1}{\left(1+x\right)^{3}}}={\frac {1}{\left(1+x\right)\left(1-x+x^{2}\right)}}={\frac {A}{1+x}}+{\frac {Bx+C}{1-x+x^{2}}}$, 則有:

${\begin{cases}A+B=0\\B+C-A=0\\A+C=1\end{cases}}$

解得$A={\frac {1}{3}}$, $B=-{\frac {1}{3}}$, $C={\frac {2}{3}}$, 所以${\frac {1}{\left(1+x\right)^{3}}}={\frac {1}{3\left(1+x\right)}}-{\frac {x-2}{3\left(1-x+x^{2}\right)}}$

${\begin{aligned}\int _{0}^{\infty }{\frac {\mathrm {d} x}{\left(1+x\right)^{3}}}&=\int _{0}^{\infty }{\frac {1}{3\left(1+x\right)}}-{\frac {x-2}{3\left(1-x+x^{2}\right)}}\mathrm {d} x\\&={\frac {1}{3}}\ln \left(x+1\right){\big |}_{0}^{\infty }-{\frac {1}{3}}\int _{0}^{\infty }{\frac {{\tfrac {1}{2}}(2x-1)-{\tfrac {3}{2}}}{x^{2}-x+1}}\mathrm {d} x\\&={\frac {1}{3}}\ln \left(x+1\right){\big |}_{0}^{\infty }-{\frac {1}{6}}\int _{0}^{\infty }{\frac {\mathrm {d} \left(x^{2}-x+1\right)}{x^{2}-x+1}}+{\frac {1}{2}}\int _{0}^{\infty }{\frac {\mathrm {d} x}{x^{2}-x+1}}\\&={\frac {1}{3}}\ln \left(x+1\right){\big |}_{0}^{\infty }-{\frac {1}{6}}\ln \left(x^{2}-x+1\right){\big |}_{0}^{\infty }+\int _{0}^{\infty }{\frac {\mathrm {d} \left(2x-1\right)}{\left(2x-1\right)^{2}+3}}\\&={\frac {1}{3}}\ln \left(x+1\right){\big |}_{0}^{\infty }-{\frac {1}{6}}\ln \left(x^{2}-x+1\right){\big |}_{0}^{\infty }+{\frac {1}{\sqrt {3}}}\arctan {\frac {2x-1}{\sqrt {3}}}{\big |}_{0}^{\infty }\\&={\frac {1}{6}}\ln {\frac {x^{2}+2x+1}{x^{2}-x+1}}{\big |}_{0}^{\infty }+{\frac {1}{\sqrt {3}}}\arctan {\frac {2x-1}{\sqrt {3}}}{\big |}_{0}^{\infty }\\&=0+{\frac {1}{\sqrt {3}}}\left(\lim _{x\to \infty }\arctan {\frac {2x-1}{\sqrt {3}}}+\arctan {\frac {1}{\sqrt {3}}}\right)={\frac {2\pi }{3{\sqrt {3}}}}\\\end{aligned}}$

二

$\int _{0}^{10}x-\lfloor x\rfloor -\sin 2\pi x\mathrm {d} x=5$

解: 注意到$f(x)=x-\lfloor x\rfloor -\sin 2\pi x$是周期為1的函數，且在$\left[0,1\right]$上為$f(x)=x-\sin 2\pi x$.

故$\int _{0}^{10}x-\lfloor x\rfloor -\sin 2\pi x\mathrm {d} x=10\int _{0}^{1}x-\sin 2\pi x\mathrm {d} x=10\times {\frac {1}{2}}=5$

三

$\int _{-1}^{1}{\sqrt {1+4x^{2}}}\mathrm {d} x={\sqrt {5}}+{\frac {1}{2}}\ln \left(2+{\sqrt {5}}\right)$

解：令$\tan \theta =2x$，則$\mathrm {d} x={\frac {\mathrm {d} \theta }{2\cos ^{2}\theta }}$. 又$f(x)={\sqrt {1+4x^{2}}}$為偶函數

${\begin{aligned}\int _{-1}^{1}{\sqrt {1+4x^{2}}}\mathrm {d} x&=2\int _{0}^{1}{\sqrt {1+4x^{2}}}\mathrm {d} x\\&=\int _{0}^{\arctan 2}{\frac {\sqrt {1+\tan ^{2}\theta }}{\cos ^{2}\theta }}\mathrm {d} \theta \\&=\int _{0}^{\arctan 2}{\frac {\mathrm {d} \theta }{\cos ^{3}\theta }}\\&=\int _{0}^{\arctan 2}{\frac {\mathrm {d} (\sin \theta )}{\left(1-\sin ^{2}\theta \right)^{2}}}\\\end{aligned}}$

令$t=\sin \theta$，由於$\sin \arctan 2={\frac {2}{\sqrt {5}}}$，故

$\int _{0}^{\arctan 2}{\frac {\mathrm {d} (\sin \theta )}{\left(1-\sin ^{2}\theta \right)^{2}}}=\int _{0}^{\tfrac {2}{\sqrt {5}}}{\frac {\mathrm {d} t}{\left(1-t^{2}\right)^{2}}}$

考慮積分$I_{n}=\int {\frac {\mathrm {d} x}{\left(x^{2}-a^{2}\right)^{n}}},\ n\geqslant 2$的遞推公式為：

$I_{n}=-{\frac {1}{a^{2}}}\left[{\frac {1}{2(n-1)}}\cdot {\frac {x}{(x^{2}-a^{2})^{n-1}}}+{\frac {2n-3}{2n-2}}I_{n-1}\right],\ n\geqslant 2$

代入得：

${\begin{aligned}\int _{0}^{\tfrac {2}{\sqrt {5}}}{\frac {\mathrm {d} t}{\left(1-t^{2}\right)^{2}}}&={\frac {t}{2(1-t^{2})}}{\bigg |}_{0}^{\tfrac {2}{\sqrt {5}}}-{\frac {1}{2}}\int _{0}^{\tfrac {2}{\sqrt {5}}}{\frac {\mathrm {d} t}{1-t^{2}}}\\&={\frac {t}{2(1-t^{2})}}{\bigg |}_{0}^{\tfrac {2}{\sqrt {5}}}-{\frac {1}{4}}\ln {\frac {1-t}{1+t}}{\bigg |}_{0}^{\tfrac {2}{\sqrt {5}}}\\&={\sqrt {5}}-{\frac {1}{4}}\ln {\frac {1-{\frac {2}{\sqrt {5}}}}{1+{\frac {2}{\sqrt {5}}}}}={\sqrt {5}}+{\frac {1}{2}}\ln \left(2+{\sqrt {5}}\right)\\\end{aligned}}$

三

一

等差數列$\{a_{n}\}$，記$S_{n}=\sum _{k=1}^{n}a_{k}$，數列$\{b_{n}\}$滿足$b_{1}=5a_{1}=5$，$a_{5}=b_{2}=9$，當$n\geqslant 3$時，$S_{n+1}>b_{n}$, 且$S_{n},\ S_{n+1}-b_{n},\ S_{n-2}$成等比數列，$n\in \mathbb {N} ^{*}$

(1) 求$\{a_{n}\}$，$\{b_{n}\}$的通項公式

(2) 求證：$\{b_{n}\}\subseteq \{a_{n}\}$

(3) 將數列$\{a_{n}\}$，$\{{\frac {1}{b_{n}b_{n+1}}}\}$的項按照以下規律交叉排列：當$n$為奇數時，$\{a_{n}\}$放在前面；當$n$為偶數時，${\frac {1}{b_{n}b_{n+1}}}$放在前面，得到一個新數列$\{c_{n}\}$：$a_{1},\ {\frac {1}{b_{1}b_{2}}},\ {\frac {1}{b_{2}b_{3}}},\ a_{2},\ a_{3},\ {\frac {1}{b_{3}b_{4}}},\ {\frac {1}{b_{4}b_{5}}},\cdots$，記$T_{n}=\sum _{k=1}^{n}c_{k}$，求$T_{n}$表達式.

解：(1) $a_{1}=1$，$a_{5}=9$，又$\{a_{n}\}$為等差數列，故$a_{n}=2n-1$，則$S_{n}=n^{2}$.

由於$S_{n},\ S_{n+1}-b_{n},\ S_{n-2}$成等比數列，得：$\left[(n+1)^{2}-b_{n}\right]^{2}=n^{2}(n-2)^{2},\ (n\geqslant 3)$，又因為$S_{n+1}=(n+1)^{2}>b_{n}$，因此整理得$b_{n}=4n+1,\ (n\geqslant 3)$. 檢驗$b_{1},\ b_{2}$符合，即$b_{n}=4n+1$

(2) $\forall n\in \mathbb {N} ^{*},\ b_{n}=4n+1=2(2n)+1$. 因為$2n\in \mathbb {N} ^{*}$，所以$\forall n\in \mathbb {N} ^{*},\ b_{n}=2(2n)+1=a_{2n}$，證畢.

(3) 不妨先羅列出數列$\{c_{n}\}$的前12項：

$a_{1},\ {\frac {1}{b_{1}b_{2}}},\ {\frac {1}{b_{2}b_{3}}},\ a_{2},\ a_{3},\ {\frac {1}{b_{3}b_{4}}},\ {\frac {1}{b_{4}b_{5}}},\ a_{4},\ a_{5},\ {\frac {1}{b_{5}b_{6}}},\ {\frac {1}{b_{6}b_{7}}},\ a_{6},\cdots$

注意到$\{c_{n}\}$中的項可以每四個一組進行劃分：

$\left(a_{1},\ {\frac {1}{b_{1}b_{2}}},\ {\frac {1}{b_{2}b_{3}}},\ a_{2}\right),\ \left(a_{3},\ {\frac {1}{b_{3}b_{4}}},\ {\frac {1}{b_{4}b_{5}}},\ a_{4}\right),\ \left(a_{5},\ {\frac {1}{b_{5}b_{6}}},\ {\frac {1}{b_{6}b_{7}}},\ a_{6}\right),\cdots$

因此分情況討論如下：

1) 當$n\equiv 1{\pmod {4}}$時，記$n=4i+1,\ i\in \mathbb {N}$

此時$\{c_{n}\}$的倒數第二組與最後一個數為：

$\cdots ,\ \left(a_{2i-1},\ {\frac {1}{b_{2i-1}b_{2i}}},\ {\frac {1}{b_{2i}b_{2i+1}}},\ a_{2i}\right),\ a_{2i+1}$

${\begin{aligned}T_{n}&=\sum _{j=1}^{2i+1}a_{j}+\sum _{m=1}^{2i}b_{m}\\&=\left(a_{1}+a_{2}+\cdots +a_{\tfrac {n+1}{2}}\right)+\left({\frac {1}{5\times 9}}+{\frac {1}{9\times 13}}+\cdots +{\frac {1}{b_{\tfrac {n-1}{2}}b_{\tfrac {n+1}{2}}}}\right)\\&=\left({\frac {n+1}{2}}\right)^{2}+{\frac {1}{4}}\left({\frac {1}{5}}-{\frac {1}{b_{\tfrac {n+1}{2}}}}\right)=\left({\frac {n+1}{2}}\right)^{2}+{\frac {1}{4}}\left({\frac {1}{5}}-{\frac {1}{2n+3}}\right)\\\end{aligned}}$

2) 當$n\equiv 2{\pmod {4}}$時，記$n=4i+2,\ i\in \mathbb {N}$

此時$\{c_{n}\}$的倒數第二組與最後兩個數為：

$\cdots ,\ \left(a_{2i-1},\ {\frac {1}{b_{2i-1}b_{2i}}},\ {\frac {1}{b_{2i}b_{2i+1}}},\ a_{2i}\right),\ a_{2i+1},\ {\frac {1}{b_{2i+1}b_{2i+2}}}$

${\begin{aligned}T_{n}&=\sum _{j=1}^{2i+1}a_{j}+\sum _{m=1}^{2i+1}b_{m}\\&=\left(a_{1}+a_{2}+\cdots +a_{\tfrac {n}{2}}\right)+\left({\frac {1}{5\times 9}}+{\frac {1}{9\times 13}}+\cdots +{\frac {1}{b_{\tfrac {n}{2}}b_{\tfrac {n+2}{2}}}}\right)\\&=\left({\frac {n}{2}}\right)^{2}+{\frac {1}{4}}\left({\frac {1}{5}}-{\frac {1}{b_{\tfrac {n+2}{2}}}}\right)=\left({\frac {n}{2}}\right)^{2}+{\frac {1}{4}}\left({\frac {1}{5}}-{\frac {1}{2n+5}}\right)\\\end{aligned}}$

3) 當$n\equiv 3{\pmod {4}}$時，記$n=4i+3,\ i\in \mathbb {N}$

此時$\{c_{n}\}$的倒數第二組與最後三個數為：

$\cdots ,\ \left(a_{2i-1},\ {\frac {1}{b_{2i-1}b_{2i}}},\ {\frac {1}{b_{2i}b_{2i+1}}},\ a_{2i}\right),\ a_{2i+1},\ {\frac {1}{b_{2i+1}b_{2i+2}}},\ {\frac {1}{b_{2i+2}b_{2i+3}}}$

${\begin{aligned}T_{n}&=\sum _{j=1}^{2i+1}a_{j}+\sum _{m=1}^{2i+2}b_{m}\\&=\left(a_{1}+a_{2}+\cdots +a_{\tfrac {n-1}{2}}\right)+\left({\frac {1}{5\times 9}}+{\frac {1}{9\times 13}}+\cdots +{\frac {1}{b_{\tfrac {n+1}{2}}b_{\tfrac {n+3}{2}}}}\right)\\&=\left({\frac {n-1}{2}}\right)^{2}+{\frac {1}{4}}\left({\frac {1}{5}}-{\frac {1}{b_{\tfrac {n+3}{2}}}}\right)=\left({\frac {n-1}{2}}\right)^{2}+{\frac {1}{4}}\left({\frac {1}{5}}-{\frac {1}{2n+7}}\right)\\\end{aligned}}$

4) 當$n\equiv 0{\pmod {4}}$時，記$n=4(i+1),\ i\in \mathbb {N}$

此時$\{c_{n}\}$的倒數第二組與最後一組為：

$\cdots ,\ \left(a_{2i-1},\ {\frac {1}{b_{2i-1}b_{2i}}},\ {\frac {1}{b_{2i}b_{2i+1}}},\ a_{2i}\right),\ \left(a_{2i+1},\ {\frac {1}{b_{2i+1}b_{2i+2}}},\ {\frac {1}{b_{2i+2}b_{2i+3}}},\ a_{2i+2}\right)$

${\begin{aligned}T_{n}&=\sum _{j=1}^{2i+2}a_{j}+\sum _{m=1}^{2i+2}b_{m}\\&=\left(a_{1}+a_{2}+\cdots +a_{\tfrac {n}{2}}\right)+\left({\frac {1}{5\times 9}}+{\frac {1}{9\times 13}}+\cdots +{\frac {1}{b_{\tfrac {n}{2}}b_{\tfrac {n+2}{2}}}}\right)\\&=\left({\frac {n}{2}}\right)^{2}+{\frac {1}{4}}\left({\frac {1}{5}}-{\frac {1}{b_{\tfrac {n+2}{2}}}}\right)=\left({\frac {n}{2}}\right)^{2}+{\frac {1}{4}}\left({\frac {1}{5}}-{\frac {1}{2n+5}}\right)\\\end{aligned}}$

所以綜上，得：

$T_{n}={\begin{cases}\left({\frac {n+1}{2}}\right)^{2}+{\frac {1}{4}}\left({\frac {1}{5}}-{\frac {1}{2n+3}}\right)&\qquad n\equiv 1{\pmod {4}}\\\left({\frac {n}{2}}\right)^{2}+{\frac {1}{4}}\left({\frac {1}{5}}-{\frac {1}{2n+5}}\right)&\qquad n\equiv 2{\pmod {4}}\\\left({\frac {n-1}{2}}\right)^{2}+{\frac {1}{4}}\left({\frac {1}{5}}-{\frac {1}{2n+7}}\right)&\qquad n\equiv 3{\pmod {4}}\\\left({\frac {n}{2}}\right)^{2}+{\frac {1}{4}}\left({\frac {1}{5}}-{\frac {1}{2n+5}}\right)&\qquad n\equiv 0{\pmod {4}}\end{cases}}$

二

橢圓$C_{1}:{\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1,\ (a>b>0)$的離心率為${\frac {1}{2}}$，過點$({\sqrt {7}},\ 0)$作橢圓$C_{1}$的兩條切線相互垂直.

(1) 求橢圓$C_{1}$的方程

(2) 在橢圓$C_{1}$上是否存在點$P$：過點$P$引拋物線$x^{2}=4y$的兩條切線$l_{1},\ l_{2}$，切點分別為$B,\ C$，且直線$BC$過點$A(1,1)$？若存在，指出這樣的點$P$有幾個（不必求出點的坐標）；若不存在，請說明理由.

解：(1) 由題，$e={\frac {1}{2}}\Rightarrow C_{1}:{\frac {x^{2}}{4c^{2}}}+{\frac {y^{2}}{3c^{2}}}=1$. 由於兩切線關於$x$軸對稱且互相垂直，故斜率為$\pm 1$. 取$l_{1}:y=-x+{\sqrt {7}}$，與橢圓方程聯立：

${\begin{cases}{\frac {x^{2}}{4c^{2}}}+{\frac {y^{2}}{3c^{2}}}=1\\y=-x+{\sqrt {7}}\\\end{cases}}\Rightarrow 7x^{2}-8{\sqrt {7}}x+28-12c^{2}=0$

故$\Delta =448-28(28-12c^{2})=0\Rightarrow c=1$，即$C_{1}:{\frac {x^{2}}{4}}+{\frac {y^{2}}{3}}=1$

(2) 顯然，直接設點$P$坐標將會十分麻煩. 不妨設$BC:y=m(x-1)+1$，與拋物線交點為$B(x_{1},{\frac {x_{1}^{2}}{4}}),\ C(x_{2},{\frac {x_{2}^{2}}{4}}),\ x_{1}<x_{2}$，切線$l_{1},\ l_{2}$的斜率分別為$k_{1},\ k_{2}$. 聯立兩方程：

${\begin{cases}x^{2}=4y\\y=m(x-1)+1\\\end{cases}}\Rightarrow x^{2}-4mx+4m-4=0\Rightarrow x_{1}+x_{2}=4m,\ x_{1}x_{2}=4m-4$

由於$k_{1}={\frac {1}{2}}x_{1}\Rightarrow l_{1}:y={\frac {1}{2}}x_{1}x-{\frac {x_{1}^{2}}{4}}$. 同理$l_{2}:y={\frac {1}{2}}x_{2}x-{\frac {x_{2}^{2}}{4}}$. 聯立$l_{1},\ l_{2}$方程

${\begin{cases}y={\frac {1}{2}}x_{1}x-{\frac {x_{1}^{2}}{4}}\\y={\frac {1}{2}}x_{2}x-{\frac {x_{2}^{2}}{4}}\\\end{cases}}\Rightarrow {\begin{cases}x={\frac {x_{1}+x_{2}}{2}}=2m\\y={\frac {1}{4}}x_{1}x_{2}=m-1\\\end{cases}}$

故得$P(2m,m-1)$. 根據題意，$P\in C_{1}$，故代入得：${\frac {4m^{2}}{4}}+{\frac {(m-1)^{2}}{3}}=1$，解得$m_{1}=-{\frac {1}{2}},\ m_{2}=1$. 故點$P$有兩個，分別為$\left(-1,-{\frac {3}{2}}\right)$與$\left(2,0\right)$