How do you find the points on the parabola #2x = y^2# that are closest to the point (3,0)?

Answer 1

Points are (2.2) and (2, -2)

Let there be any point (x,y) on this parabola. The distance 's' of this point from point (3,0) is given by #s^2= (x-3)^2 +y^2#. Differentiate both sides w.r.t x
#2s(ds)/dx= 2(x-3) + 2ydy/dx#. For minimum distance #(ds)/dx# =0, hence (x-3)+#ydy/dx=0#, Or y#dy/dx#= 3-x
Differentiating the equation #y^2#=2x with respect to x, it would be y#dy/dx=1#

It is thus 3-x=1, x=2, and then y=2, -2. The nearest points are (2,2) and (2,-2)

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Answer 2

An alternative starts the same as bp's solution:

Let #(x,y)# be any point on this parabola. The distance 's' of this point from point #(3,0)# is given by #s^2= (x-3)^2 +y^2#.
Note that, since #(x,y)# is on the graph, we must have #y^2=2x#, so
#s^2= (x-3)^2 +2x#

Our job now is to minimize the function:

#f(x) = (x-3)^2 +2x = x^2-4x+9#

(It should be clear that we can minimize the distance by minimizing the square of the distance.)

To minimize, differentiate, find and test critical numbers.

#f'(x) = 2x-4#, so the critical number is #2#
#f''(2) = 2# is positive, so #f(2)# is a minimum.
The question asks for points on the graph, so we finish by finding points on the graph at which #x=2#
#2(2)=y^2# has solutions #y=+-2#
The points are: #(2,2)# and #(2,-2)#
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Answer 3

To find the points on the parabola (2x = y^2) that are closest to the point ((3,0)), you can use the method of Lagrange multipliers. The distance between two points ((x, y)) and ((3, 0)) is given by the distance formula. By minimizing this distance, you can find the points on the parabola closest to ((3,0)).

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Answer 4

To find the points on the parabola (2x = y^2) that are closest to the point ((3, 0)), we need to minimize the distance between the given point and the points on the parabola.

The distance between two points ((x_1, y_1)) and ((x_2, y_2)) is given by the distance formula:

[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}]

For any point ((x, y)) on the parabola (2x = y^2), the distance (d) from ((3, 0)) can be expressed as:

[d = \sqrt{(x - 3)^2 + y^2}]

We want to minimize (d). Since (d) is a positive quantity, minimizing (d^2) is equivalent and easier. Thus, we minimize (d^2) instead:

[d^2 = (x - 3)^2 + y^2]

Substitute (2x = y^2) into the equation:

[d^2 = (x - 3)^2 + (2x)^2]

[d^2 = (x - 3)^2 + 4x^2]

Now, we take the derivative of (d^2) with respect to (x) and set it equal to zero to find the critical points:

[\frac{d}{dx}(d^2) = \frac{d}{dx}[(x - 3)^2 + 4x^2]]

[0 = 2(x - 3) + 8x]

[0 = 2x - 6 + 8x]

[0 = 10x - 6]

[x = \frac{6}{10} = \frac{3}{5}]

Substitute (x = \frac{3}{5}) into (2x = y^2) to find (y):

[2\left(\frac{3}{5}\right) = y^2]

[y^2 = \frac{6}{5}]

[y = \pm \sqrt{\frac{6}{5}}]

Thus, the points on the parabola (2x = y^2) that are closest to the point ((3, 0)) are (\left(\frac{3}{5}, \sqrt{\frac{6}{5}}\right)) and (\left(\frac{3}{5}, -\sqrt{\frac{6}{5}}\right)).

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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.

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.

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.

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.

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