Find two points on an ellpise that have horizontal tangent

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SUMMARY

The discussion focuses on finding two points on the ellipse defined by the equation 5x² - 6xy + 5y² = 16 where the tangent is horizontal. The derivative is calculated as (-10x + 6y) / (-6x + 10y) = 0, leading to the condition -10x + 6y = 0, which simplifies to x = (3/5)y. Substituting this back into the original equation yields incorrect values for y, indicating an error in the back substitution process.

PREREQUISITES
  • Understanding of implicit differentiation
  • Familiarity with conic sections, specifically ellipses
  • Knowledge of solving systems of equations
  • Ability to perform back substitution in algebraic equations
NEXT STEPS
  • Review implicit differentiation techniques in calculus
  • Study the properties and equations of ellipses
  • Practice solving systems of equations involving multiple variables
  • Learn about back substitution methods in algebra
USEFUL FOR

Students studying calculus, particularly those focusing on conic sections and implicit differentiation, as well as educators looking for examples of horizontal tangents on ellipses.

zeion
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Homework Statement



The equation 5x^2 - 6xy + 5y^2 = 16 represents an ellipse.

Determine two points on the ellipse at which the tangent is horizontal.


Homework Equations





The Attempt at a Solution



I find the derivative of the equation:

(-10x + 6y) / (-6x + 10y) = 0 iff -10x + 6y = 0, so x = (3/5)y.
Then I sub x back into original equation and get y, I get y = +/- 1/(sqrt(2)), which is wrong.

What did I do wrong?
 
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zeion said:
so x = (3/5)y.

This is correct...

zeion said:
Then I sub x back into original equation and get y, I get y = +/- 1/(sqrt(2)), which is wrong.

What did I do wrong?

Something with the back substitution into the original equation.

ehild
 
Got it, thanks.
 

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