题目

Problem

8. The point $P(2p^2, 4p)$ lies on the parabola with equation $y^2 = 8x$

(a) Show that the point $Q\Big(\dfrac{2}{p^2}, -\dfrac{4}{p}\Big)$ , where $p \neq 0$ , lies on the parabola.

(1)

(b) Show that the chord $PQ$ passes through the focus of the parabola.

(4)

The tangent to the parabola at $P$ and the tangent to the parabola at $Q$ meet at the point $R$

(8)

题目中文翻译

点 $P(2p^2, 4p)$ 在抛物线 $y^2 = 8x$ 上。

(a) 证明点 $Q\Big(\dfrac{2}{p^2}, -\dfrac{4}{p}\Big)$ （其中 $p \neq 0$ ）在抛物线上。

(1)

(b) 证明弦 $PQ$ 经过抛物线的焦点。

(4)

抛物线在 $P$ 处的切线和抛物线在 $Q$ 处的切线相交于点 $R$ 。

(8)

解答

(a)

解法一

思路

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把 $Q$ 的横、纵坐标代入 $y^2=8x$ 。若等式两边相同，就证明了 $Q$ 在抛物线上。

答题过程

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At $Q\big(2/p^2,-4/p\big)$ ,

\begin{align*} y^2=&\,\Big(-\frac4p\Big)^2\\ =&\,\frac{16}{p^2}, \end{align*}

and

\begin{align*} 8x=&\,8\Big(\frac2{p^2}\Big)\\ =&\,\frac{16}{p^2}. \end{align*}

Therefore $y^2=8x$ , so $Q$ lies on the parabola.

(b)

解法一

思路

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由 $y^2=4ax$ 可知焦点为 $(2,0)$ 。利用 $P,Q$ 写出弦的方程，再代入焦点坐标验证。另检查 $p=\pm1$ 时的竖直弦，避免在求斜率时除以零。

答题过程

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Since $y^2=8x=4(2)x$ , the focus is

F=(2,0).

First suppose $p^2\ne1$ . The gradient of $PQ$ is

\begin{align*} &\,m_{PQ}\\ =&\,\frac{4p-(-4/p)}{2p^2-2/p^2}\\ =&\,\frac{4p^3+4p}{2p^4-2}\\ =&\,\frac{4p(p^2+1)}{2(p^2-1)(p^2+1)}\\ =&\,\frac{2p}{p^2-1}. \end{align*}

Here $p\ne0$ and $p^2+1>0$ , so the cancellations are valid. Using $P$ ,

\begin{align*} &\,y-4p\\ =&\,\frac{2p}{p^2-1}(x-2p^2). \end{align*}

Multiplying by $p^2-1$ gives

\begin{align*} &\,(p^2-1)(y-4p)\\ =&\,2p(x-2p^2). \end{align*}

Expanding,

\begin{align*} &\,(p^2-1)y-4p^3+4p\\ =&\,2px-4p^3. \end{align*}

Cancelling $-4p^3$ and rearranging gives

\begin{align*} &\,(p^2-1)y\\ =&\,2px-4p. \end{align*}

Substituting $F=(2,0)$ , the two sides are

(p^2-1)(0)=0

and

2p(2)-4p=0.

Thus $F$ lies on $PQ$ .

If $p=\pm1$ , both $P$ and $Q$ have $x$ -coordinate $2$ , so $PQ$ is the vertical line $x=2$ . The cross-multiplied equation above also reduces to $x=2$ . This line contains $F$ as well.

Therefore the chord $PQ$ passes through the focus for every allowed value of $p$ .

解法二

思路

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比较 $PF$ 与 $QF$ 的斜率。若两条过焦点的直线斜率相同，则 $P,Q,F$ 共线； $p=\pm1$ 时单独检查竖直线。

答题过程

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The focus is $F=(2,0)$ . For $p^2\ne1$ ,

\begin{align*} &\,m_{PF}\\ =&\,\frac{4p-0}{2p^2-2}\\ =&\,\frac{2p}{p^2-1}. \end{align*}

Also,

\begin{align*} &\,m_{QF}\\ =&\,\frac{-4/p-0}{2/p^2-2}\\ =&\,\frac{-4p}{2-2p^2}\\ =&\,\frac{2p}{p^2-1}. \end{align*}

Hence $m_{PF}=m_{QF}$ . As both lines pass through $F$ , the points $P,Q,F$ are collinear.

For $p=\pm1$ , $P,Q,F$ all have $x$ -coordinate $2$ , so they are again collinear. Thus $PQ$ passes through the focus.

解法三

思路

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在经过 $Q$ 的水平线上作两个辅助点：一个位于焦点正下方或正上方，另一个位于 $P$ 的正下方或正上方。比较两个直角三角形的竖直边和水平边之比，证明三角形相似，再推出 $Q,F,P$ 共线。长度用绝对值表示，因此不必假设 $p$ 为正或 $p^2>1$ 。

答题过程

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The focus is $F=(2,0)$ . First take $p^2\ne1$ . Define

X=\Big(2,-\frac4p\Big)

and

Y=\Big(2p^2,-\frac4p\Big).

The points $Q,X,Y$ lie on the same horizontal line. The segments $XF$ and $YP$ are vertical, so $\triangle QXF$ and $\triangle QYP$ are right-angled at $X$ and $Y$ respectively.

Their vertical side lengths are

XF=\frac4{|p|}

and

\begin{align*} &\,YP\\ =&\,\Big|4p+\frac4p\Big|\\ =&\,\frac{4(p^2+1)}{|p|}. \end{align*}

Consequently,

\begin{align*} &\,\frac{XF}{YP}\\ =&\,\frac{4/|p|}{4(p^2+1)/|p|}\\ =&\,\frac1{p^2+1}. \end{align*}

The corresponding horizontal side lengths are

\begin{align*} &\,QX\\ =&\,\Big|2-\frac2{p^2}\Big|\\ =&\,\frac{2|p^2-1|}{p^2} \end{align*}

and

\begin{align*} &\,QY\\ =&\,\Big|2p^2-\frac2{p^2}\Big|\\ =&\,\frac{2|p^4-1|}{p^2}\\ =&\,\frac{2|p^2-1|(p^2+1)}{p^2}. \end{align*}

Cancelling the positive common factor $2|p^2-1|/p^2$ gives

\frac{QX}{QY}=\frac1{p^2+1}.

Thus

\frac{XF}{YP}=\frac{QX}{QY}.

The corresponding legs of the right triangles are proportional, so

\triangle QXF\sim\triangle QYP.

Also, $X$ and $Y$ lie on the same horizontal ray from $Q$ , and $XF$ and $YP$ point in the same vertical direction. The equal angles at $Q$ therefore show that $QF$ and $QP$ are the same line. Hence $P,Q,F$ are collinear.

For $p=\pm1$ , the chord is the vertical line $x=2$ , which contains $F$ . Thus $PQ$ passes through the focus in this case too.

(c)

解法一

思路

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对抛物线隐函数求导，分别求出 $P,Q$ 处的切线斜率。写出两条切线方程后联立，先解得 $x=-2$ ，再代回求纵坐标。

答题过程

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Differentiating $y^2=8x$ with respect to $x$ gives

2y\frac{\mathrm{d}y}{\mathrm{d}x}=8,

\frac{\mathrm{d}y}{\mathrm{d}x}=\frac4y.

At $P(2p^2,4p)$ , the tangent gradient is

\frac4{4p}=\frac1p.

The tangent at $P$ is therefore

\begin{align*} &\,y-4p\\ =&\,\frac1p(x-2p^2). \end{align*}

Expanding and collecting terms,

\begin{align*} y=&\,\frac xp-2p+4p\\ =&\,\frac xp+2p. \end{align*}

At $Q\big(2/p^2,-4/p\big)$ , the tangent gradient is

\frac4{-4/p}=-p.

Its tangent is

\begin{align*} &\,y+\frac4p\\ =&\,-p\Big(x-\frac2{p^2}\Big). \end{align*}

Hence

\begin{align*} y=&\,-px+\frac2p-\frac4p\\ =&\,-px-\frac2p. \end{align*}

At their point of intersection,

\begin{align*} &\,\frac xp+2p\\ =&\,-px-\frac2p. \end{align*}

Multiplying by the non-zero parameter $p$ and rearranging,

\begin{align*} x+2p^2=&\,-p^2x-2,\\ (1+p^2)x=&\,-2-2p^2,\\ (1+p^2)x=&\,-2(1+p^2). \end{align*}

Since $1+p^2>0$ , division gives

x=-2.

Substituting into the tangent at $P$ ,

\begin{align*} y=&\,\frac{-2}{p}+2p\\ =&\,2p-\frac2p. \end{align*}

Hence

\boxed{R\Big(-2,\,2p-\frac2p\Big)}.

解法二

思路

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用参数形式求切线方程。 $P$ 对应参数 $p$ ， $Q$ 对应参数 $-1/p$ 。联立两条切线时，改为先消去 $x$ 求 $y$ ，再回代求 $x$ 。

答题过程

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Parametrise the parabola by

x=2t^2,

y=4t.

For $t\ne0$ ,

\frac{\mathrm{d}x}{\mathrm{d}t}=4t

and

\frac{\mathrm{d}y}{\mathrm{d}t}=4.

Thus the tangent gradient is

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac4{4t}\\ =&\,\frac1t. \end{align*}

The tangent at parameter $t$ is

\begin{align*} &\,y-4t\\ =&\,\frac1t(x-2t^2). \end{align*}

Multiplying by $t$ ,

ty-4t^2=x-2t^2,

ty=x+2t^2.

At $P$ , $t=p$ , giving

py=x+2p^2.

At $Q$ , $t=-1/p$ , since

2t^2=\frac2{p^2},

4t=-\frac4p.

Therefore its tangent satisfies

-\frac yp=x+\frac2{p^2}.

Multiplying by $-p$ gives

y=-px-\frac2p.

From the tangent at $P$ ,

x=py-2p^2.

Substitute this into the tangent at $Q$ :

\begin{align*} &\,y\\ =&\,-p(py-2p^2)-\frac2p\\ =&\,-p^2y+2p^3-\frac2p. \end{align*}

Consequently,

\begin{align*} &\,(1+p^2)y\\ =&\,2p^3-\frac2p\\ =&\,\frac{2p^4-2}{p}\\ =&\,\frac{2(p^2-1)(p^2+1)}p. \end{align*}

Since $1+p^2>0$ , cancel this factor to obtain

\begin{align*} y=&\,\frac{2(p^2-1)}p\\ =&\,2p-\frac2p. \end{align*}

Substituting back,

\begin{align*} &\,x\\ =&\,p\Big(2p-\frac2p\Big)-2p^2\\ =&\,2p^2-2-2p^2\\ =&\,-2. \end{align*}

Hence

\boxed{R\Big(-2,\,2p-\frac2p\Big)}.

FP1 2023 May Q8

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

解法二

思路

答题过程

解法三

思路

答题过程

(c)

解法一

思路

答题过程

解法二

思路

答题过程