Find a function with given condition

Click For Summary
SUMMARY

The discussion focuses on finding a curve that passes through the point A(2,0) such that the triangle formed by the tangent at an arbitrary point M, the y-axis, and the secant \overline{OM} is isosceles. The function is represented as y=k(x-2) where k is a non-zero constant, leading to two cases based on the sign of k. The key insight is that the distance from point M to the y-axis must equal the distance from the origin to the tangent's intersection with the y-axis, resulting in a first-order differential equation that the function f must satisfy, with the initial condition f(2) = 0.

PREREQUISITES
  • Understanding of first-order differential equations
  • Knowledge of tangent lines and their properties
  • Familiarity with the concept of isosceles triangles in coordinate geometry
  • Basic skills in curve sketching and analysis
NEXT STEPS
  • Study the properties of first-order differential equations
  • Learn about tangent lines and their equations in calculus
  • Explore the geometric properties of isosceles triangles in the Cartesian plane
  • Investigate curve sketching techniques for polynomial functions
USEFUL FOR

Students in calculus, mathematicians interested in differential equations, and educators teaching geometric properties of curves and triangles.

gruba
Messages
203
Reaction score
1

Homework Statement


Find a curve that passes through point [itex]A(2,0)[/itex] such that the triangle which is defined with a tangent at arbitrary point [itex]M[/itex], axis [itex]Oy[/itex] and secant [itex]\overline{OM}[/itex] is isosceles. [itex]\overline{OM}[/itex] is the base side of a triangle.

2. The attempt at a solution
Function passes through point [itex]A(2,0)\Rightarrow y=k(x-2),k\neq 0.[/itex]
There are two cases, for [itex]k<0[/itex] and for [itex]k>0[/itex].

What is the relation between points [itex]A[/itex] and [itex]M[/itex] because [itex]M[/itex] is not defined?

In which quadrant the point [itex]M[/itex] should be?
 
Physics news on Phys.org
M is an arbitrary point in the curve; its coordinates are (x, f(x)).

I think the assertion is that the distance along the tangent from M to the y-axis is equal to the distance from the origin to the intersection of the tangent and the y-axis. That allows you to write down a first-order differential equation which f must satisfy, and your initial condition comes from the constraint that A lies on the curve (ie f(2) = 0).
 
  • Like
Likes   Reactions: BvU

Similar threads

  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 39 ·
2
Replies
39
Views
6K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 0 ·
Replies
0
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K