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FP1 2022 Jan Q7

A Level / Edexcel / FP1

IAL 2022 Jan Paper · Question 7

题目

Problem

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

The rectangular hyperbola HH has equation xy=36xy = 36

The point P(4,9)P(4, 9) lies on HH

(a) Show, using calculus, that the normal to HH at PP has equation

4x−9y+65=04x - 9y + 65 = 0
(4)

The normal to HH at PP crosses HH again at the point QQ

(b) Determine an equation for the tangent to HH at QQ, giving your answer in the form y=mx+cy = mx + c where mm and cc are rational constants.

(5)
题目中文翻译

本题必须展示所有解题步骤。 完全依赖计算器技术的解答不可接受。

等轴双曲线 HH 的方程为 xy=36xy = 36。

点 P(4,9)P(4, 9) 在 HH 上。

(a) 使用微积分证明：HH 在点 PP 处的法线方程为

4x−9y+65=04x - 9y + 65 = 0
(4)

HH 在点 PP 处的法线再次与 HH 相交于点 QQ。

(b) 确定 HH 在点 QQ 处的切线方程，答案以 y=mx+cy = mx + c 的形式表示，其中 mm 和 cc 为有理常数。

(5)

解答

(a)

解法一

思路

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对 xy=36xy=36 隐函数求导，代入 P(4,9)P(4,9) 求切线斜率，再取负倒数得到法线斜率。用点斜式写出法线并逐步整理到题目给定形式。

答题过程

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Differentiating xy=36xy=36 implicitly with respect to xx gives

xdydx+y=0,x\frac{\mathrm{d}y}{\mathrm{d}x}+y=0,

so

dydx= −yx= −36/xx= −36x2.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,-\frac yx\\ =&\,-\frac{36/x}{x}\\ =&\,-\frac{36}{x^2}. \end{align*}

At P(4,9)P(4,9), the tangent gradient is

mT=−94.m_T=-\frac94.

Hence the normal gradient is

mN= −1mT= −1−9/4= 49.\begin{align*} m_N=&\,-\frac1{m_T}\\ =&\,-\frac1{-9/4}\\ =&\,\frac49. \end{align*}

The normal at PP is therefore

y−9=49(x−4).y-9=\frac49(x-4).

Multiplying by 99 and rearranging,

 9y−81= 4x−16, 0= 4x−9y+81−16= 4x−9y+65.\begin{align*} &\,9y-81\\ =&\,4x-16,\\[4mm] &\,0\\ =&\,4x-9y+81-16\\ =&\,4x-9y+65. \end{align*}

Thus the required equation is

4x−9y+65=0.\boxed{4x-9y+65=0}.

解法二

思路

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先把双曲线写成 y=36x−1y=36x^{-1}，使用幂函数求导法则直接求切线斜率。随后取负倒数得到法线斜率，再代入点 PP 写出法线。

答题过程

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Since xy=36xy=36,

y=36x−1.y=36x^{-1}.

Differentiating with respect to xx,

dydx= 36(−1)x−2= −36x2.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,36(-1)x^{-2}\\ =&\,-\frac{36}{x^2}. \end{align*}

At P(4,9)P(4,9),

mT= −3642= −3616= −94.\begin{align*} m_T =&\,-\frac{36}{4^2}\\ =&\,-\frac{36}{16}\\ =&\,-\frac94. \end{align*}

The normal gradient is

mN= −1−9/4= 49.\begin{align*} m_N =&\,-\frac1{-9/4}\\ =&\,\frac49. \end{align*}

Using the point PP,

y−9=49(x−4).y-9=\frac49(x-4).

Hence

 9y−81= 4x−16, 0= 4x−9y+81−16= 4x−9y+65.\begin{align*} &\,9y-81\\ =&\,4x-16,\\[4mm] &\,0\\ =&\,4x-9y+81-16\\ =&\,4x-9y+65. \end{align*}

Therefore, the required normal is

4x−9y+65=0.\boxed{4x-9y+65=0}.

(b)

解法一

思路

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先由法线方程把 yy 表示成 xx，代入 xy=36xy=36 求两个交点的横坐标；排除已知点 PP 后得到 QQ。再在 QQ 处求切线斜率，并用点斜式写成题目要求的形式。

答题过程

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From the normal equation,

y=4x+659.y=\frac{4x+65}{9}.

At an intersection with HH,

x(4x+659)=36.x\big(\frac{4x+65}{9}\big)=36.

Thus

4x2+65x= 324,4x2+65x−324= 0.\begin{align*} 4x^2+65x=&\,324,\\ 4x^2+65x-324=&\,0. \end{align*}

Factorising the left-hand side,

 4x2+65x−324= 4x2+81x−16x−324= x(4x+81)−4(4x+81)= (x−4)(4x+81).\begin{align*} &\,4x^2+65x-324\\ =&\,4x^2+81x-16x-324\\ =&\,x(4x+81)-4(4x+81)\\ =&\,(x-4)(4x+81). \end{align*}

Thus

(x−4)(4x+81)= 0.\begin{align*} (x-4)(4x+81)=&\,0. \end{align*}

The root x=4x=4 corresponds to PP. Hence at QQ,

xQ=−814,x_Q=-\frac{81}{4},

and

yQ= 36−81/4= 36×(−481)= −14481= −169.\begin{align*} y_Q =&\,\frac{36}{-81/4}\\ =&\,36\times\big(-\frac4{81}\big)\\ =&\,-\frac{144}{81}\\ =&\,-\frac{16}{9}. \end{align*}

Therefore,

Q(−814,−169).Q\big(-\frac{81}{4},-\frac{16}{9}\big).

Since y=36x−1y=36x^{-1},

dydx=−36x−2.\frac{\mathrm{d}y}{\mathrm{d}x}=-36x^{-2}.

The tangent gradient at QQ is

m= −36(−81/4)2= −366561/16= −36×166561= −5766561= −64729.\begin{align*} m =&\,-\frac{36}{(-81/4)^2}\\ =&\,-\frac{36}{6561/16}\\ =&\,-36\times\frac{16}{6561}\\ =&\,-\frac{576}{6561}\\ =&\,-\frac{64}{729}. \end{align*}

Hence the tangent is

 y+169= −64729(x+814).\begin{align*} &\,y+\frac{16}{9}\\ =&\,-\frac{64}{729} \big(x+\frac{81}{4}\big). \end{align*}

The constant term on the right is

−64729×814= −169.\begin{align*} -\frac{64}{729}\times\frac{81}{4} =&\,-\frac{16}{9}. \end{align*}

Hence

 y+169= −64729x−169,y= −64729x −169−169= −64729x−329.\begin{align*} &\,y+\frac{16}{9}\\ =&\,-\frac{64}{729}x-\frac{16}{9},\\[4mm] y =&\,-\frac{64}{729}x\\ &\,\hspace{2pt}-\frac{16}{9}-\frac{16}{9}\\ =&\,\boxed{-\frac{64}{729}x-\frac{32}{9}}. \end{align*}

解法二

思路

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也可以改为消去 xx。由法线方程把 xx 表示成 yy，先求 QQ 的纵坐标，再由 xy=36xy=36 求横坐标；得到 QQ 后的切线步骤相同。

答题过程

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From 4x−9y+65=04x-9y+65=0,

x=9y−654.x=\frac{9y-65}{4}.

Substituting into xy=36xy=36 gives

y(9y−654)=36.y\big(\frac{9y-65}{4}\big)=36.

Hence

9y2−65y= 144,9y2−65y−144= 0.\begin{align*} 9y^2-65y=&\,144,\\ 9y^2-65y-144=&\,0. \end{align*}

Factorising the left-hand side,

 9y2−65y−144= 9y2+16y−81y−144= y(9y+16)−9(9y+16)= (y−9)(9y+16).\begin{align*} &\,9y^2-65y-144\\ =&\,9y^2+16y-81y-144\\ =&\,y(9y+16)-9(9y+16)\\ =&\,(y-9)(9y+16). \end{align*}

Thus

(y−9)(9y+16)= 0.\begin{align*} (y-9)(9y+16)=&\,0. \end{align*}

The root y=9y=9 corresponds to PP, so

yQ=−169y_Q=-\frac{16}{9}

and

xQ= 36−16/9= 36×(−916)= −32416= −814.\begin{align*} x_Q =&\,\frac{36}{-16/9}\\ =&\,36\times\big(-\frac9{16}\big)\\ =&\,-\frac{324}{16}\\ =&\,-\frac{81}{4}. \end{align*}

Thus Q=(−81/4,−16/9)Q=(-81/4,-16/9). Using the derivative found in part (a),

m= −yQxQ= −−16/9−81/4= 169×(−481)= −64729.\begin{align*} m =&\,-\frac{y_Q}{x_Q}\\ =&\,-\frac{-16/9}{-81/4}\\ =&\,\frac{16}{9}\times\big(-\frac4{81}\big)\\ =&\,-\frac{64}{729}. \end{align*}

Therefore,

 y+169= −64729(x+814)= −64729x−169.\begin{align*} &\,y+\frac{16}{9}\\ =&\,-\frac{64}{729} \big(x+\frac{81}{4}\big)\\ =&\,-\frac{64}{729}x-\frac{16}{9}. \end{align*}

Subtracting 16/916/9 from both sides gives

y= −64729x −169−169= −64729x−329.\begin{align*} y =&\,-\frac{64}{729}x\\ &\,\hspace{2pt}-\frac{16}{9}-\frac{16}{9}\\ =&\,\boxed{-\frac{64}{729}x-\frac{32}{9}}. \end{align*}