# The midpoint of the line segment cut from the tangent line by coordinate axes P . ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 3.1, Problem 70E

(a)

To determine

## To show: The midpoint of the line segment cut from the tangent line by coordinate axes P.

Expert Solution

### Explanation of Solution

Given:

The equation of the hyperbola is xy=c (or)y=cx.

Derivative rules:

(1) Power Rule: ddx(xn)=nxn1

(2) Constant multiple rule: ddx(cf)=cddx(f)

Formula Used:

The equation of the tangent line at (x1,y1) is, yy1=m(xx1). (1)

where, m is the slope of the tangent line at (x1,y1) and m=dydx|x=x1.

Proof:

Obtain the slope of the tangent line to the hyperbola at P.

The derivative of hyperbola y is dydx as follows,

dydx=ddx(cx)=ddx(cx1)

Apply the constant multiple rule (2) and the power rule (1),

dydx=cddx(x1)=c(1x11)=cx2=cx2

Thus, the derivative of y is cx2.

Therefore, the slope of the tangent to the hyperbola is dydx=cx2.

Since the hyperbola is y=cx and choose the point P is (a,ca), the slope of the tangent line to the hyperbola at P is ca2.

Substitute (a,ca) for (x1,y1) and ca2 for m in equation (1),

yca=ca2(xa)y=ca2x+ca+cay=ca2x+2ca

Thus, the equation of tangent line at P is y=ca2x+2ca.

Substitute 0 for x in y=ca2x+2ca, the y-intercept is 2ca  and the point (0,2ca).

Substitute 0 for y in y=ca2x+2ca, the x-intercept is computed as follows.

ca2x+2ca=0ca2x=2cax=2ca×a2cx=2a

Thus, the x-intercept of line y=ca2x+2ca is x=2a and the point (2a,0).

The midpoint of the line segment joining the points (2a,0) and (0,2ca) is computed as follows,

(x,y)=(2a+02,0+2ca2)=(a,ca)

Therefore, it can be concluded that the midpoint of the line segment cut from the tangent line by coordinate axes is P.

(b)

To determine

Expert Solution

### Explanation of Solution

Proof:

From part (a), the x-intercept of the tangent line is 2a and the y-intercept of the tangent line is 2ca.

Here, the base b is x-intercept of the tangent line and the height h is y-intercept of the tangent line.

The area of the triangle bounded by axes and tangent line is computed as follows.

12bh=12(2a)(2ca)=2c

It is obvious that the area of the triangle is a constant and it is independent of x and y.

Therefore, it can be concluded that the triangle formed by the tangent line and the coordinate axes always has the same area.

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