   Chapter 3.P, Problem 9P

Chapter
Section
Textbook Problem

# If P ( a , a 2 ) is any point on the parabola y = x 2 , except for the origin, let Q be the point where the normal line at P intersects the parabola again (see the figure).(a) Show that the y-coordinate of Q is smallest when a = 1 / 2 (b) Show that the line segment PQ has the shortest possible length when a = 1 / 2 . s

To determine

a)

To show:

That the y-coordinate of Q is smallest when a=12

Explanation

1) Concept:

First calculate the formula of the normal line then using point slope form find intersection point Q with the parabola and then take derivative of that point with respect to a.

2) Formula:

i. Quadratic formula for ax2+bx+c=0 is

x=-b±b2-4ac2a

ii. Point slope form-y-y0=dydx(x-x0)

3) Given:

y=x2

4) Calculation:

First calculate the formula of the normal line:

Given the parabola

y=x2

Differentiate it.

dydx=2x

At point (a, a2) the slope of the tangent line is 2a.

Therefore the slope of the normal line is -12a.

Using the point slope form:

y-a2=-12a(x-a)

y-a2+a2=-12ax-a+a2

Simplify.

y=-x2a+12+a2

Now to find the intersection point Q with the parabola;

Put y=x2 in point slope form.

x2=-x2a+12+a2

Subtract -x2a+12+a2 from both sides.

x2--x2a+12+a2=-x2a+12+a2--x2a+12+a2

Simplify.

x2+x2a-12-a2=0

x=-12a±12a2-4·1·-12-a22·1

Simplify.

x=-12a±14a2+2+4a22

x=-14a±1214a2+2+4a2

Factor out 14a2 from the square root term

To determine

b)

To show:

That the line segment PQ has the shortest possible length when a=12..

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