题目

Problem

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

(a) Show that the point $Q\left(\dfrac{2}{p^2}, -\dfrac{4}{p}\right)$ , 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 $Q\left(\dfrac{2}{p^2}, -\dfrac{4}{p}\right)$ （其中 $p \neq 0$ ）在抛物线上。

(b) 证明弦 $PQ$ 经过抛物线的焦点。

抛物线在 $P$ 处的切线和抛物线在 $Q$ 处的切线相交于点 $R$ 。

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 $Q\left(2/p^2,-4/p\right)$ ,

y^2=\left(-\frac4p\right)^2=\frac{16}{p^2},

and

8x=8\left(\frac2{p^2}\right)=\frac{16}{p^2}.

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；将方程写成交叉相乘后的形式，也能涵盖弦为竖直线的 $p=\pm1$ 情形。

答题过程

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

F=(2,0).

The gradient of $PQ$ , when $p^2\ne1$ , is

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

Hence the equation of the chord, in cross-multiplied form, is

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

This equation also gives the vertical chords $x=2$ when $p=\pm1$ . Substituting $F=(2,0)$ gives

(p^2-1)(0)=2p(2)-4p=0.

Therefore $F$ lies on $PQ$ , so the chord $PQ$ passes through the focus.

(c)

解法一

思路

展开

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 $1/p$ . Its equation is therefore

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

At $Q\left(2/p^2,-4/p\right)$ , the tangent gradient is $-p$ . Its equation is

\begin{align*} y+\frac4p=&\,-p\left(x-\frac2{p^2}\right),\\ y=&\,-px-\frac2p. \end{align*}

At their point of intersection,

\frac{x}{p}+2p=-px-\frac2p.

Multiplying by $p$ and rearranging,

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

Substituting into the tangent at $P$ ,

y=\frac{-2}{p}+2p=2p-\frac2p.

Hence

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

IAL 2023 May FP1 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

(c)

解法一

思路

答题过程