题目

Problem

The curve $C_1$ has equation

\begin{align*} y=x^2+kx-9 \end{align*}

and the curve $C_2$ has equation

\begin{align*} y=-3x^2-5x+k \end{align*}

where $k$ is a constant.

Given that $C_1$ and $C_2$ meet at a single point $P$ ,

(a) show that

\begin{align*} k^2+26k+169=0. \end{align*}

(3)

(b) Hence find the coordinates of $P$ .

(3)

解答

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两曲线交点满足两个 $y$ 相等。整理后得到关于 $x$ 的二次方程。它们只交于一个点，所以这个二次方程有重根，即判别式等于 $0$ 。

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At intersections,

\begin{align*} x^2+kx-9=-3x^2-5x+k. \end{align*}

Rearrange:

\begin{align*} 4x^2+(k+5)x-(9+k)=0. \end{align*}

Since the curves meet at a single point, this quadratic has equal roots, so

\begin{align*} b^2-4ac=0. \end{align*}

Thus

\begin{align*} (k+5)^2-4(4)(-9-k)=&\,0\\ k^2+10k+25+144+16k=&\,0\\ k^2+26k+169=&\,0. \end{align*}

This is the required result.

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先解 (a) 的方程求 $k$ ，再代回交点方程求 $x$ 。因为只有一个交点，得到的二次方程会是完全平方。

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From part (a),

\begin{align*} k^2+26k+169=&\,0\\ (k+13)^2=&\,0\\ k=&\,-13. \end{align*}

Substitute $k=-13$ into the intersection equation:

\begin{align*} 4x^2+(k+5)x-(9+k)=&\,0\\ 4x^2-8x+4=&\,0\\ x^2-2x+1=&\,0\\ (x-1)^2=&\,0. \end{align*}

\begin{align*} x=1. \end{align*}

Use $C_1$ :

\begin{align*} y =&\,1^2+(-13)(1)-9\\ =&\,-21. \end{align*}

Therefore

\begin{align*} P=(1,-21). \end{align*}