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Which point on the parabola #y=x^2# is nearest to (1,0)?

Answer 1

Approximately #(0.58975, 0.34781)#

I'm just going to solve this by the first method that comes to me, rather than trying to use any special geometric properties of parabolas.

If #(x, y)# is a point on the parabola, then the distance between #(x, y)# and #(1, 0)# is:

#sqrt((x-1)^2+(y-0)^2) = sqrt(x^4+x^2-2x+1)#

To minimize this, we want to minimize #f(x) = x^4+x^2-2x+1#

The minimum will occur at a zero of:

#f'(x) = 4x^3+2x-2 = 2(2x^3+x-1)#

graph{2x^3+x-1 [-10, 10, -5, 5]}

Using Cardano's method, find

#x = root(3)(1/4 + sqrt(87)/36) + root(3)(1/4 - sqrt(87)/36) ~= 0.58975#

#y = x^2 ~= 0.34781#

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Answer 2

Scarlett Allison

To find the point on the parabola y = x^2 nearest to (1,0), we need to minimize the distance between this point and (1,0). The point on the parabola can be represented as (x, x^2). Using the distance formula, the distance between (1,0) and (x, x^2) is sqrt((x - 1)^2 + (x^2 - 0)^2). To minimize this distance, differentiate the expression with respect to x, set it equal to zero, and solve for x. This yields x = 1/2. Therefore, the point on the parabola nearest to (1,0) is (1/2, 1/4).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.