Consider line segments which are tangent to a point on the right half (x>0) of the curve y = x^2 + 1 and connect the tangent point to the x-axis. If the tangent point is close to the y-axis, the line segment is long. If the tangent point is far from the y-axis, the line segment is also very long. Which tangent point has the shortest line segment?

(Suppose C is a positive number. What point on the curve has first coordinate equal to C? What is the slope of the tangent line at that point? Find the x-intercept of the resulting line. Compute the distance between the point on the curve and the x-intecept, and find the minimum of the square of that distance (minimizing the square of a positive quantity gets the same answer as minimizing the quantity, and here we get rid of a square root).)

following the hints suggested:

let the point be (c,c^2 + 1)
dy/dx = 2x
so at (c,c^2+1) the slope of the tangent is 2c

let the tangent equation be y = mx + b
y = 2cx + b
for our point,
c^2 + 1 = 2c(c) + b
b = 1-c^2

so the tangent equation is
y = 2cx + 1-c^2
at the x-intercept,
0 = 2cx + 1-c^2
x = (c^2 - 1)/(2c)

then using the distance formula

D^2 = (c^2+1)^2 + (c - (c^2 - 1)/(2c))^2

Ok, I will now expand this. At first I thought to differenctiate using quotient rule for the last term, but it looked rather messy

D^2 = c^4 + 2c^2 + 1 + c^2 - c^2 + 1 + (c^4-2c^2+1)/(4c^2)
= c^4 + 2c^2 + 2 + (1/4)c^2 - 1/2 + (1/4)c^-2

2D(dD/dc) = 4c^3 + 4c + c/2 - 1/(2c^3) = 0 for a max/min of D
8c^6 + 9c^4 - 1 = 0

getting really messy....
let a = c^2
solve 8a^3 + 9a^2 - 1 = 0
a=-1 works !!!!!!
(a+1)(8a^2 + a - 1) = 0
if a=-1, c^=-1 ---> no solution
8a^2 + a - 1 = 0
a = (-1 ± √33)/16 = .2965 or a negative
c^2 = .2965
c = .5145

Please, please check my arithmetic and algebra, the method is right!

Thanks so much, but then do you plug in c back into the first point and tangent line and x-intercept to get the x and y coordinates and tangent line equation and x-intercept, respectively?

Can you explain to me how you found out the distance formula. Because how did you find out D^2 = c^4 + 2c^2 + 1 + c^2 - c^2 + 1 + (c^4-2c^2+1)/(4c^2)? I understand the first part with c^4 + 2c^2 + 1, but I don't understand how you got the second part.

And isn't (c^4-2c^2+1)/(4c^2) supposed to be (c^4+2c^2+1)/(4c^2)?

To find the tangent point with the shortest line segment, we need to minimize the square of the distance between the point on the curve and the x-intercept of the resulting line. Let's break down the steps to find this point:

1. Start by finding the first coordinate of the tangent point on the curve. Let's call this coordinate C.

2. Substitute the value of C into the equation of the curve, y = x^2 + 1, to find the corresponding y-value.

3. To find the slope of the tangent line at the point (C, y), we need to find the derivative of the curve with respect to x.

Take the derivative of y = x^2 + 1 with respect to x:
dy/dx = 2x

Evaluate the derivative at x = C to find the slope of the tangent line at that point.

4. Use the slope-intercept form of a line, y = mx + b, where m is the slope and b is the y-intercept, to find the equation of the tangent line.

The point-slope form of a line is: y - y1 = m(x - x1), where (x1, y1) is the point on the line.

We have the point (C, y) and the slope dy/dx, so substitute these values into the point-slope form to get the equation of the tangent line.

5. To find the x-intercept of the tangent line, set y to zero in the equation of the tangent line and solve for x.

This will give you the x-coordinate of the x-intercept.

6. Compute the distance between the point on the curve (C, y) and the x-intercept of the tangent line.

The distance formula is: distance = √((x2 - x1)^2 + (y2 - y1)^2), where (x1, y1) and (x2, y2) are the coordinates of two points.

Substitute the coordinates of the point on the curve and the x-intercept into the distance formula.

7. Square the distance obtained in step 6 to get rid of the square root.

8. Repeat steps 1-7 for different values of C on the right half of the curve.

9. Compare the squared distances calculated in step 7 for each value of C. The tangent point with the smallest squared distance will have the shortest line segment.

By following these steps, you can find the tangent point with the shortest line segment on the right half of the curve y = x^2 + 1.