The the points on the curve x2=8y
nearest to the origin is

To find the point on the curve \( x^2 = 8y \) that is nearest to the origin \((0,0)\), we need to minimize the distance between a point \((x, y)\) on the curve and the origin. 

The distance \( D \) from a point \((x, y)\) to the origin is given by:

\[
D = \sqrt{x^2 + y^2}
\]

Since \( y \) is a function of \( x \) (from the curve equation), we can express \( D \) in terms of \( x \) only:

1. The curve is given as \( x^2 = 8y \), so \( y = \frac{x^2}{8} \).
2. Substitute \( y = \frac{x^2}{8} \) into the distance formula:

\[
D = \sqrt{x^2 + \left(\frac{x^2}{8}\right)^2}
\]

To minimize \( D \), we can minimize \( D^2 \) instead, to avoid dealing with the square root:

\[
D^2 = x^2 + \frac{x^4}{64}
\]

Next, differentiate \( D^2 \) with respect to \( x \) and set the derivative equal to zero to find the critical points:

\[
\frac{d(D^2)}{dx} = 2x + \frac{4x^3}{64} = 2x + \frac{x^3}{16}
\]

Set the derivative equal to zero:

\[
2x + \frac{x^3}{16} = 0
\]

Factor the equation:

\[
x\left(2 + \frac{x^2}{16}\right) = 0
\]

This gives two cases:

1. \( x = 0 \)
2. \( 2 + \frac{x^2}{16} = 0 \) which simplifies to \( x^2 = -32 \) (not possible since \( x^2 \) cannot be negative).

So, \( x = 0 \) is the only solution. 

Substitute \( x = 0 \) into the curve equation \( x^2 = 8y \) to find \( y \):

\[
0^2 = 8y \implies y = 0
\]

Thus, the point on the curve nearest to the origin is \((0, 0)\).

Would you like more details or have any questions about this solution? Here are some related questions you might find interesting:

1. How do you find the distance from a point to a curve in general?
2. What is the method of Lagrange multipliers, and how is it used to solve constrained optimization problems?
3. Can the distance formula be generalized to higher dimensions?
4. What are the properties of the parabola \( x^2 = 8y \)?
5. How does the concept of minimizing distance relate to gradient descent methods?

**Tip:** In optimization problems, minimizing or maximizing a function often involves finding the derivative and solving for critical points where the derivative is zero.

Learn how to find the nearest point on the curve x^2 = 8y to the origin using the distance formula and optimization techniques. This problem involves geometry, specifically finding points on a parabolic curve closest to a given point, suitable for advanced high school students.

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