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IAL 2024 Jan FP1 Q7

A Level / Edexcel / FP1

IAL 2024 Jan Paper · Question 7

题目

Problem

The parabola CC has equation y2=43xy^2 = \dfrac{4}{3}x

The point P(13t2,23t)P\left(\dfrac{1}{3}t^2, \dfrac{2}{3}t\right), where t≠0t \neq 0, lies on CC.

(a) Use calculus to show that the normal to CC at PP has equation

3tx+3y=t3+2t3tx + 3y = t^3 + 2t

(3)

The normal to CC at the point where t=9t = 9 meets CC again at the point QQ.

(b) Determine the exact coordinates of QQ.

(4)
题目中文翻译

抛物线 CC 的方程为 y2=43xy^2 = \dfrac{4}{3}x。

点 P(13t2,23t)P\left(\dfrac{1}{3}t^2, \dfrac{2}{3}t\right)（其中 t≠0t \neq 0）在 CC 上。

(a) 使用微积分证明：CC 在点 PP 处的法线方程为 3tx+3y=t3+2t3tx + 3y = t^3 + 2t

CC 在 t=9t = 9 处的法线再次与 CC 相交于点 QQ。

(b) 确定 QQ 的精确坐标。

解答

(a)

解法一

思路

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把 xx、yy 都视为参数 tt 的函数，用参数方程求导得到切线斜率；法线斜率是其负倒数。再将点 PP 代入点斜式并整理。

答题过程

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Since

x=13t2andy=23t,x=\frac13t^2 \quad\text{and}\quad y=\frac23t, dxdt=23tanddydt=23.\frac{\mathrm{d}x}{\mathrm{d}t}=\frac23t \quad\text{and}\quad \frac{\mathrm{d}y}{\mathrm{d}t}=\frac23.

Therefore,

dydx=dy/dtdx/dt=1t.\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t} =\frac1t.

As t≠0t\ne0, the gradient of the normal is −t-t. Hence its equation at PP is

y−23t=−t(x−13t2).y-\frac23t=-t\left(x-\frac13t^2\right).

Multiplying by 3 and rearranging gives

3tx+3y=t3+2t.\boxed{3tx+3y=t^3+2t}.

(b)

解法一

思路

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令 t=9t=9 得到法线方程，再用抛物线方程将 xx 写成 yy 的二次式。所得二次方程的一个根对应已知交点，另一个根便对应 QQ。

答题过程

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When t=9t=9, the equation of the normal is

27x+3y=747,27x+3y=747,

so

9x+y=249.9x+y=249.

Since QQ also lies on CC,

x=34y2.x=\frac34y^2.

Substituting into the normal gives

9(34y2)+y= 249,27y2+4y−996= 0,(y−6)(27y+166)= 0.\begin{align*} 9\left(\frac34y^2\right)+y =&\,249,\\ 27y^2+4y-996 =&\,0,\\ (y-6)(27y+166) =&\,0. \end{align*}

The root y=6y=6 corresponds to the original point where t=9t=9. Therefore, at QQ,

y=−16627.y=-\frac{166}{27}.

Thus

x=34(−16627)2=6889243.x=\frac34\left(-\frac{166}{27}\right)^2 =\frac{6889}{243}.

Hence

Q(6889243,−16627).\boxed{Q\left(\frac{6889}{243},-\frac{166}{27}\right)}.

解法二

思路

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直接以另一个参数表示法线与抛物线的交点。将抛物线参数式代入 t=9t=9 时的法线，二次方程的两个参数根分别对应原点 PP 与第二交点 QQ。

答题过程

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Let the parameter of a point where the normal meets CC be uu. Then

x=13u2,y=23u.x=\frac13u^2, \qquad y=\frac23u.

Substituting these into 9x+y=2499x+y=249 gives

3u2+23u= 249,9u2+2u−747= 0,(u−9)(9u+83)= 0.\begin{align*} 3u^2+\frac23u =&\,249,\\ 9u^2+2u-747 =&\,0,\\ (u-9)(9u+83) =&\,0. \end{align*}

The root u=9u=9 corresponds to the original point. Hence the parameter of QQ is

u=−839.u=-\frac{83}{9}.

Therefore,

xQ= 13(−839)2=6889243,yQ= 23(−839)=−16627.\begin{align*} x_Q=&\,\frac13\left(-\frac{83}{9}\right)^2 =\frac{6889}{243},\\ y_Q=&\,\frac23\left(-\frac{83}{9}\right) =-\frac{166}{27}. \end{align*}

Thus

Q(6889243,−16627).\boxed{Q\left(\frac{6889}{243},-\frac{166}{27}\right)}.