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2021-01-11 16:48发布

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如下图，C 点是以「交点」的方式求得的，并不知其确切坐标，欲作原点(0,0) 关于 点 C 对称的点Q，也即是在 x 轴上找一点Q，令OC=CQ，求问如何引用『线短OC之长』？ ![](https...

如下图，C 点是以「交点」的方式求得的，并不知其确切坐标，欲作原点(0,0) 关于 点 C 对称的点Q，也即是在 x 轴上找一点Q，令OC=CQ，求问如何引用『线短OC之长』？ ![](https://wenda.latexstudio.net/data/attach/210111/eTvz2Rsn.png) ```lua $$\begin{tikzpicture}[scale=0.7] \draw[-latex] (-7,0) -- (12,0) node [above]{$x$}; \draw[-latex] (0,-7) -- (0,8) node [right]{$y$}; \draw plot[samples=400,domain=0:15](\x,{2*sqrt(\x)}) ; \draw[name path=a,yscale=-1] plot[samples=400,domain=0:10](\x,{2*sqrt(\x)}) node[right]{$y^2=4{x}$}; \coordinate(A) at(1,2); \coordinate(B) at(14.66,7.66); \draw[name path=f](A)--(B); \coordinate (Q) at ($(A)!-7cm!90:(B)$); \draw [name path=b](Q)--(A); \draw[name path=o] (-7,0) -- (12,0) ; \draw[name path=t] (0,-7) -- (0,8); \fill [name intersections={of=a and b, by={D}}] (D) circle (4pt) node at (3.6,-3) {$D$}; \draw[name path=c,domain=-7:4] plot(\x,{(-2*\x-6.69)/3.66}); \draw[name path=d,domain=-7:15] plot(\x,{(2*\x+29.31)/7.66}) ; \fill [name intersections={of=c and d, by={P}}] (P) circle (4pt) node[above] {$P$}; \fill [name intersections={of=o and b, by={C}}] (C) circle (4pt) node at (2,.6) {$C$}; \fill (A) circle (4pt) node at (1,2.5) {$A$}; \fill (B) circle (4pt) node at (14,8) {$B$}; \fill (0,0) circle (4pt) node at (-0.5,-0.5) {$O$}; \draw[name path=g](P)--(A); \coordinate (W) at ($(A)!-5.5cm!(B)$); \draw [name path=w](A)--(W); \fill [name intersections={of=o and w, by={I}}] (I) circle (4pt) node[below] {$I$}; \end{tikzpicture}$$ ```
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U9430
2021-01-14 11:46
```c \begin{tikzpicture} \def\pointradius{1.2pt} \draw[-latex,name path=x] (-6,0) -- (9,0) node [above]{$x$};%x轴 \draw[-latex] (0,-6) -- (0,6) node [right]{$y$};%y轴 \coordinate (O) at(0,0); \fill (O) circle [radius=\pointradius] node[below left]{$O$}; \draw[name path=a]%曲线 plot[variable=\t,samples=400,domain=-6:6]({(\t)^2/4},\t) coordinate (B); \coordinate (A) at({(2)^2/4},2); \fill (B) circle [radius=\pointradius] node[above]{$B$} (A) circle [radius=\pointradius] node[above]{$A$}; \node[right]at([yscale=-1]B){$y^2=4x$}; \path [overlay,name path=AD](A)--($(A)!1!-90:(B)$); \filldraw [name intersections={of=AD and a, by=D}] (A)--(D) circle [radius=\pointradius] node [below] {$D$}; \filldraw [name intersections={of=AD and x, by=C}] (C) circle [radius=\pointradius] node [above right] {$C$}; \fill ($(O)!2!(C)$) circle [radius=\pointradius] node[above]{$Q$} coordinate (Q); %计算切线的倾斜角度，用 y=2*sqrt(x), y=-2*sqrt(x) 的导数 \path (B); \pgfgetlastxy{\macrox}{\macroy} \pgfmathparse{atan(1/sqrt(\macrox/28.45274))}%1cm = 28.45274pt \let\tBangle\pgfmathresult \path [overlay,name path=BP] (B)--++(\tBangle+180:50); \path (D); \pgfgetlastxy{\macrox}{\macroy} \pgfmathparse{atan(-1/sqrt(\macrox/28.45274))}%1cm = 28.45274pt \let\tDangle\pgfmathresult \path[overlay,name path=DP] (D)--++(\tDangle+180:50); \filldraw [name intersections={of=DP and BP, by=P}] (D)--(P) (B)--(P) circle [radius=\pointradius] node [above] {$P$}; \path[overlay,name path=AB] (A)--($(B)!10!(A)$); \filldraw [name intersections={of=x and AB, by=I}] (B)--(I) circle [radius=\pointradius] node [above] {$I$}; \end{tikzpicture} ``` 效果是 ![](https://wenda.latexstudio.net/data/attach/210114/4DZQqQZG.png)

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