Math Problem Statement

Le plan est muni d'un repère orthonormé (O, ī, j̄). Soit D la droite d'équation y = 2x + 3. Soit M un point de la droite D et a son abscisse. 1)a) Montrer que OM^2 = 5(a + 6/5)^2 + 9/5. b) En déduire la valeur de a pour laquelle la distance OM est minimale. 2) Tracer la droite D et construire le point M0 pour lequel la distance OM est minimale.

Solution

Let's analyze and work through this problem step-by-step.

Given:

  1. A line DD with the equation y=2x+3y = 2x + 3.
  2. A point MM on this line with abscissa aa.

We need to solve the following:

Part 1a: Show that OM2=5(a+65)2+95OM^2 = 5 \left( a + \frac{6}{5} \right)^2 + \frac{9}{5}

Solution Outline:

  1. Since MM is on the line DD, with abscissa aa, its coordinates are: M(a,2a+3)M(a, 2a + 3)
  2. The origin OO has coordinates (0,0)(0, 0).
  3. The square of the distance OMOM is given by: OM2=(a0)2+(2a+30)2OM^2 = (a - 0)^2 + (2a + 3 - 0)^2
  4. Substitute (a,2a+3)(a, 2a + 3) into this expression and expand it to obtain a form that matches the required result.

I'll work through these calculations now.

Part 1b: Find the value of aa that minimizes OMOM

Once we have the expression for OM2OM^2 in the required form, we will find the minimum by differentiating with respect to aa and setting the derivative to zero.

Part 2: Graph and Construct Point M0M_0

For this part, we'll draw the line DD and locate the point M0M_0 at the minimum distance from OO.

Let me complete the calculations for Part 1a first.The expanded form of OM2OM^2 is:

OM2=5a2+12a+9OM^2 = 5a^2 + 12a + 9

To express this in the required form, OM2=5(a+65)2+95OM^2 = 5 \left( a + \frac{6}{5} \right)^2 + \frac{9}{5}, we can complete the square.

Let's proceed with completing the square for this expression.After completing the square, the expression becomes:

OM2=5(a+65)2+95OM^2 = 5 \left( a + \frac{6}{5} \right)^2 + \frac{9}{5}

This matches the required form.

Part 1b: Find the value of aa that minimizes OMOM

To minimize OMOM, we set the derivative of OM2OM^2 with respect to aa to zero. Let's calculate this next.The value of aa that minimizes OMOM is:

a=65a = -\frac{6}{5}

Summary of Results

  1. We showed that OM2=5(a+65)2+95OM^2 = 5 \left( a + \frac{6}{5} \right)^2 + \frac{9}{5}.
  2. The value of aa that minimizes the distance OMOM is a=65a = -\frac{6}{5}.

For part 2, you can now plot the line DD and locate the point M0M_0 with coordinates (65,2×65+3)\left(-\frac{6}{5}, 2 \times -\frac{6}{5} + 3 \right).

Would you like any further details or assistance with plotting, or have any additional questions?


Related Questions

  1. How do we verify the coordinates of point M0M_0 on the line DD?
  2. What is the minimum distance OMOM when a=65a = -\frac{6}{5}?
  3. How can we interpret the completed square form of OM2OM^2 geometrically?
  4. Why does setting the derivative of OM2OM^2 to zero yield the minimum distance?
  5. How would the result change if the slope of the line DD were different?

Tip

Completing the square is a useful technique for converting quadratic expressions into a form that easily reveals minimum or maximum values.

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Distance Formula
Minimization
Algebraic Manipulation
Completing the Square

Formulas

Distance formula between two points: d^2 = (x2 - x1)^2 + (y2 - y1)^2
Completing the square to simplify quadratic expressions

Theorems

Concept of perpendicular distance from a point to a line

Suitable Grade Level

Grades 10-12