What Are the Tangent Lines to an Ellipse Passing Through a Given Point?

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SUMMARY

The discussion focuses on finding the equations of the tangent lines to the ellipse defined by the equation x² + 9y² = 81 that pass through the point (27, 3). The initial attempt identified one tangent line as horizontal (y = 3) but incorrectly assumed that (27, 3) lies on the ellipse. The correct approach involves determining the slopes of the tangent lines at points on the ellipse and ensuring they match the slope of the line connecting (27, 3) to those points. This requires using the derivative of the ellipse equation and solving for the correct tangent points.

PREREQUISITES
  • Understanding of ellipse equations and their properties
  • Knowledge of derivatives and implicit differentiation
  • Ability to solve systems of equations
  • Familiarity with slope calculations in coordinate geometry
NEXT STEPS
  • Study implicit differentiation techniques for conic sections
  • Learn how to derive tangent lines to ellipses
  • Explore graphical methods for visualizing conic sections
  • Practice solving systems of equations involving conic sections
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Students studying calculus, particularly those focusing on conic sections, as well as educators seeking to clarify concepts related to tangent lines and ellipses.

General_Sax
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Homework Statement


Find the equations of both the tangent lines to the ellipse x2 + 9y2 = 81 that pass through the point (27, 3).
One is horizontal the other is not.


Homework Equations





The Attempt at a Solution


horizontal, easy: y = 3



x^2+9y^2=81
derivative:

2x + 18yy` = 0
y`= -x/9y
at the point (27,3) the slope will be -1.

y-3 = -(x-27)

y= -x + 30

this solution is wrong according to my online assignment program, but I can't for the life of me see why.
 
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General_Sax said:

Homework Statement


Find the equations of both the tangent lines to the ellipse x2 + 9y2 = 81 that pass through the point (27, 3).
One is horizontal the other is not.


Homework Equations





The Attempt at a Solution


horizontal, easy: y = 3



x^2+9y^2=81
derivative:

2x + 18yy` = 0
y`= -x/9y
at the point (27,3) the slope will be -1.

y-3 = -(x-27)

y= -x + 30

this solution is wrong according to my online assignment program, but I can't for the life of me see why.
You are assuming that the point (27, 3) is on the ellipse when you use your formula for y', but this is not a point on the ellipse. I would approach this problem by sketching a graph of the ellipse, and drawing a line from the given point to that it is tangent to the ellipse at some unknown point (x0, y0). The slope at the tangent to the ellipse at this point has to be equal to what your derivative says, and also has to be equal to the slope of the line between this point and the point (27, 3).
 

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