# The points on the parabola y = 1 − x 2 at which the tangent line cuts from first quadrant the triangle with the smallest area. ### 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 4, Problem 6P
To determine

## To find: The points on the parabola y=1−x2 at which the tangent line cuts from first quadrant the triangle with the smallest area.

Expert Solution

The point on the parabola is (13,23)

### Explanation of Solution

Let the points on the curve be (u,v) .

As the points lie on the parabola y=1x2 write v=1u2 .

Slope of the tangent line is, dydx=2x .

Slope of the tangent line at point (u,v) it is 2u .

Write the equation of the tangent line by using slope-point form.

Thus, the equation of the tangent line is, yv=2u(xu) .

Find x-intercepts and y-intercepts of the tangent.

To find x-intercept, put y=0 .

Thus, the x-intercept is u+v2u .

Similarly to find y-intercept, put x=0 .

Thus, the y-intercept is v+2u2 .

Now area of the triangle formed is given by,

A=12xinterceptyintercept=12(u+v2u)(v+2u2)=(2u2+v)24u

Substitute value of v as v=1u2 .

A=(2u2+v)24u=(2u2+1u2)24u=(1+u2)24u

Differentiate A with respect to u.

16u2(1+u2)(1+u2)2416u2=016u2(1+u2)(1+u2)24=016u2(1+u2)=(1+u2)2416u2=(1+u2)4

Simplify further as follows.

16u2=4(1+u2)12u2=4u2=412u=±13

As the point is in first quadrant u=13 .

Substitute to get value of v.

v=1u2=1(13)2=113=23

Hence, for the values u=13 and v=23 the tangent line to the curve y=1x2 cuts the first quadrant in a triangle with smallest area.

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