# The angle between the parabola and the line segment FP is equal to the angle between the parabola and the line y = y 1 and the given parabola. That is, α = β . ### 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, Problem 18P
To determine

## To show: The angle between the parabola and the line segment FP is equal to the angle between the parabola and the line y=y1 and the given parabola. That is, α=β.

Expert Solution

### Explanation of Solution

Proof:

The equation of the parabola is, y2=4px.

Let P(x1,y1) be a point on the parabola with focus F(p,0).

Let α be an angle between parabola and the line segment FP and β be the angle between the horizontal line y=y1 and parabola.

Obtain the slope of the tangent line at P(x1,y1) is computed as follows,

Differentiate y2=4px with respect to x,

ddx(y2)=ddx(4px)2ydydx=4pddx(x)dydx=2py

Thus, the slope of the tangent at P(x1,y1) is ,m1=2py1.

The slope of the line passing through P(x1,y1) and F(p,0) is computed as follows,

m2=y10x1p=y1x1p

The angle between the tangent line at P and the line FP is α,

tanα=m2m11+m1m2

Substitute m1=4p2y1 and m2=y1x1p,

tanα=y1x1p4p2y11+(y1x1p)(4p2y1)=2y124p(x1p)(x1p)(2y1)(x1p)(2y1)+y14p(x1p)(2y1)=2y124px1+4p22y1x12py1+4py1=2y124px1+4p22y1x1+2py1

Since y12=4px1,

tanα=2(4px1)4px1+4p22y1x1+2py1=4px1+4p22y1x1+2py1=4p(x1+p)2y1(x1+p)=2py1

Since the tangent line at P(x1,y1) is 2py1 and the value tanβ=2py1,

tanα=tanβ

In 0<α,β<π2,

α=β

Hence the required result is proved.

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