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直线

y=1

与双曲线

y=\frac{1}{x}

相交于点

A_{1}

, 与双曲线

y=\frac{2}{x}

相交于点

B_{1}

, 直...

Mar 9, 2024

直线

y=1

与双曲线

y=\frac{1}{x}

相交于点

A_{1}

, 与双曲线

y=\frac{2}{x}

相交于点

B_{1}

, 直线

y=2

与双曲线

y=\frac{1}{x}

相交于点

A_{2}

, 与双曲线

y=\frac{2}{x}

相交于点

B_{2}

, 则四边形

\mathrm{A}_{1} \mathrm{~B}_{1} \mathrm{~B}_{2} \mathrm{~A}_{2}

的面积为 __; 直线

y=\mathrm{n}

与双曲线

\mathrm{y}=\frac{1}{x}

相交于点

\mathrm{A}_{\mathrm{n}}

, 与双曲线

y=\frac{2}{x}

相交于点

B_{\mathrm{n}}

, 直线

y=\mathrm{n}+1

与双曲线

y=\frac{1}{x}

相交于点

\mathrm{A}_{\mathrm{a}+1}

, 与双曲线

y=\frac{2}{X}

相交于点

B_{n+1}

, 则四边形

A_{n} B_{n} B_{n+1} A_{n+1}

的面积为

Generated Graph

Solution by Steps

step 1

To find the area of the quadrilateral

A_1B_1B_2A_2

, we need to find the coordinates of the points

A_1

B_1

A_2

, and

B_2

step 2

From the asksia-ll calculator result, the intersection point

A_1

of the line

y=1

and the hyperbola

y=\frac{1}{x}

is at

x=1

. Thus,

A_1

has coordinates

(1,1)

step 3

The intersection point

B_1

of the line

y=1

and the hyperbola

y=\frac{2}{x}

is at

x=2

. Thus,

B_1

has coordinates

(2,1)

step 4

The intersection point

A_2

of the line

y=2

and the hyperbola

y=\frac{1}{x}

is at

x=\frac{1}{2}

. Thus,

A_2

has coordinates

(\frac{1}{2},2)

step 5

The intersection point

B_2

of the line

y=2

and the hyperbola

y=\frac{2}{x}

is at

x=1

. Thus,

B_2

has coordinates

(1,2)

step 6

The quadrilateral

A_1B_1B_2A_2

is a trapezoid with bases

A_1A_2

and

B_1B_2

, and the height is the distance between these bases, which is

B_1B_2 - A_1A_2 = 2 - 1 = 1

step 7

The length of base

A_1A_2

is the difference in the x-coordinates of

A_2

and

A_1

, which is

\frac{1}{2} - 1 = -\frac{1}{2}

step 8

The length of base

B_1B_2

is the difference in the x-coordinates of

B_2

and

B_1

, which is

1 - 2 = -1

step 9

The area of the trapezoid is given by the formula

\frac{1}{2} \times (base_1 + base_2) \times height

step 10

Substituting the lengths of the bases and the height, the area is

\frac{1}{2} \times (-\frac{1}{2} + (-1)) \times 1 = \frac{1}{2} \times (-\frac{3}{2}) \times 1 = -\frac{3}{4}

. Since area cannot be negative, we take the absolute value, which is

\frac{3}{4}

Answer

\frac{3}{4}

Key Concept

Area of a Trapezoid

Explanation

The area of a trapezoid is calculated using the formula

\frac{1}{2} \times (base_1 + base_2) \times height

, where

base_1

and

base_2

are the lengths of the two parallel sides, and

height

is the distance between them.

Solution by Steps

step 1

To find the area of the quadrilateral

A_nB_nB_{n+1}A_{n+1}

, we need to find the coordinates of the points

A_n

B_n

A_{n+1}

, and

B_{n+1}

for a general

n

step 2

The intersection point

A_n

of the line

y=n

and the hyperbola

y=\frac{1}{x}

is at

x=\frac{1}{n}

. Thus,

A_n

has coordinates

(\frac{1}{n},n)

step 3

The intersection point

B_n

of the line

y=n

and the hyperbola

y=\frac{2}{x}

is at

x=\frac{2}{n}

. Thus,

B_n

has coordinates

(\frac{2}{n},n)

step 4

The intersection point

A_{n+1}

of the line

y=n+1

and the hyperbola

y=\frac{1}{x}

is at

x=\frac{1}{n+1}

. Thus,

A_{n+1}

has coordinates

(\frac{1}{n+1},n+1)

step 5

The intersection point

B_{n+1}

of the line

y=n+1

and the hyperbola

y=\frac{2}{x}

is at

x=\frac{2}{n+1}

. Thus,

B_{n+1}

has coordinates

(\frac{2}{n+1},n+1)

step 6

The quadrilateral

A_nB_nB_{n+1}A_{n+1}

is a trapezoid with bases

A_nA_{n+1}

and

B_nB_{n+1}

, and the height is the distance between these bases, which is

B_nB_{n+1} - A_nA_{n+1} = n+1 - n = 1

step 7

The length of base

A_nA_{n+1}

is the difference in the x-coordinates of

A_{n+1}

and

A_n

, which is

\frac{1}{n+1} - \frac{1}{n}

step 8

The length of base

B_nB_{n+1}

is the difference in the x-coordinates of

B_{n+1}

and

B_n

, which is

\frac{2}{n+1} - \frac{2}{n}

step 9

The area of the trapezoid is given by the formula

\frac{1}{2} \times (base_1 + base_2) \times height

step 10

Substituting the lengths of the bases and the height, the area is

\frac{1}{2} \times (\frac{1}{n+1} - \frac{1}{n} + \frac{2}{n+1} - \frac{2}{n}) \times 1

step 11

Simplifying the expression for the area, we get

\frac{1}{2} \times (\frac{1-2n+n}{n(n+1)} + \frac{2-4n+2n}{n(n+1)})

step 12

Further simplifying, we find the area to be

\frac{1}{2} \times (\frac{3}{n(n+1)})

Answer

\frac{3}{2n(n+1)}

Key Concept

Area of a Trapezoid with Variable Bases

Explanation

The area of a trapezoid with variable bases can be found by substituting the expressions for the bases and the height into the trapezoid area formula. The bases are determined by the x-coordinates of the intersection points of the lines and hyperbolas for a given

n