下表给出了两个变量 $x$ 和 $y$ 的实验值，它们通过形如 $\frac{a}{y} = \frac{b}{x} + 1$ 的方程相联系，其中 $a$ 和 $b$ 为常数。

$x$	1.25	2	2.5	5
$y$	1	1.51	1.82	3.08

发现 $\frac{1}{x}$ 的值相当小。现提出另一个以 $x$ 为横轴的直线图。

1. Calculate the values of 1/x and 1/y for each data point.

For $x = 1.25, 2, 2.5, 5$:
$\frac{1}{x} = 0.8, 0.5, 0.4, 0.2$

For $y = 1, 1.51, 1.82, 3.08$:
$\frac{1}{y} = 1, 0.662, 0.549, 0.325$

2. Plot the points and draw the line of best fit.

Plot the points $(0.8, 1)$, $(0.5, 0.662)$, $(0.4, 0.549)$, and $(0.2, 0.325)$ on the grid and draw a straight line passing through or close to all points.

1. Rearrange the given equation into a linear form $Y = mX + c$ where $Y = \frac{1}{y}$ and $X = \frac{1}{x}$.

$\frac{a}{y} = \frac{b}{x} + 1$

Divide by $a$:
$\frac{1}{y} = \frac{b}{a}\left(\frac{1}{x}\right) + \frac{1}{a}$

2. Use the $Y$-intercept from the graph to find $a$.

From the linear form, the $Y$-intercept is $\frac{1}{a}$.
Using the points $(0.2, 0.325)$ and $(0.8, 1)$ to find the equation of the line:
Gradient $m = \frac{1 - 0.325}{0.8 - 0.2} = \frac{0.675}{0.6} = 1.125$
$Y$-intercept $c = 1 - 1.125(0.8) = 0.1$

$\frac{1}{a} = 0.1$
$a = 10$

3. Use the gradient from the graph to find $b$.

From the linear form, the gradient is $\frac{b}{a}$.
$\frac{b}{a} = 1.125$
$\frac{b}{10} = 1.125$
$b = 11.25$

1. Rearrange the given equation to form a linear equation with $x$ as the independent variable

Given $\frac{a}{y} = \frac{b}{x} + 1$.
Multiply the entire equation by $x$ to make $x$ the horizontal axis variable:
$\frac{ax}{y} = b + x$
$a\left(\frac{x}{y}\right) = x + b$
$\frac{x}{y} = \frac{1}{a}x + \frac{b}{a}$

1. Express the equation in the new linear form $Y = mX + c$ where $X = x$

Starting with $\frac{a}{y} = \frac{b}{x} + 1$, multiply throughout by $x$:
$\frac{ax}{y} = b + x$
$\frac{x}{y} = \frac{1}{a}x + \frac{b}{a}$

2. Explain how to find $a$ and $b$ from the gradient and intercept

Comparing $\frac{x}{y} = \frac{1}{a}x + \frac{b}{a}$ with $Y = mX + c$:
The gradient $m = \frac{1}{a}$, so $a = \frac{1}{m}$.
The vertical intercept $c = \frac{b}{a}$, so $b = ac$.