Intersection of a parabola with another curve

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SUMMARY

The discussion focuses on finding a curve that intersects a parabola defined by the equation y=kx² at right angles. The user derived the slope of the tangent to the parabola as -1/(2kx) and integrated to obtain the equation y + (1/(2k))ln(|x|) = C. However, the user encountered an issue with only one intersection point when graphing the two functions, prompting a review of their logic and integration process.

PREREQUISITES
  • Understanding of calculus, specifically derivatives and integration
  • Familiarity with the concept of slopes and perpendicular lines
  • Knowledge of logarithmic functions and their properties
  • Graphing techniques for visualizing functions and their intersections
NEXT STEPS
  • Explore the properties of parabolas and their derivatives
  • Study the concept of implicit differentiation in calculus
  • Learn about the behavior of logarithmic functions in relation to polynomial functions
  • Investigate methods for finding multiple intersections between curves
USEFUL FOR

Students studying calculus, mathematicians interested in curve analysis, and educators seeking to understand the intersection of polynomial and logarithmic functions.

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Homework Statement


For a any parabola with the equation [tex]y=kx^{2}[/tex]
I'm trying to find a curve that intersect every point of the parabola at right angles.

Homework Equations



For a perpendicular intersection the slope is [tex]-\frac{1}{m}[/tex]

The Attempt at a Solution



I took the derivative and then took the negative reciprocal of the derivative.

[tex]\frac{dy}{dx} = -\frac{1}{2kx}[/tex]

Then I isolated the variables on different sides and then integrated. I ended up with:

[tex]y+ \frac{1}{2k}*ln(|x|) = 0[/tex]

My problem is when I graph the two functions there is only one intersection and I was wondering if there was any flaws in my logic I used to reach my answer.
 
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Don't forget that you get a constant +C when you integrate.

so

[tex] y+ \frac{1}{2k}*ln(|x|) = 0 [/tex]

would actually be

[tex] y+ \frac{1}{2k}*ln(|x|) = C [/tex]
 

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