Equation of Tangent for Ellipse ax^2+by^2=1

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SUMMARY

The tangent to the ellipse defined by the equation ax2 + by2 = 1 at the point (h, k) is given by the equation ahx + bky = 1. Furthermore, the chord of contact of tangents from the point (m, n) to the ellipse can be expressed as amx + bny = 1. The discussion emphasizes the need to deduce the relationship between the tangent and the chord of contact using the properties of the ellipse and the gradients of the lines involved.

PREREQUISITES
  • Understanding of ellipse equations, specifically ax2 + by2 = 1.
  • Knowledge of tangent lines and their equations in coordinate geometry.
  • Familiarity with the concept of the chord of contact in relation to conic sections.
  • Ability to manipulate algebraic expressions and equations involving slopes.
NEXT STEPS
  • Study the derivation of the tangent line equation for ellipses in detail.
  • Learn about the properties of chords of contact for different conic sections.
  • Explore the geometric interpretation of gradients and slopes in relation to tangents.
  • Investigate the fixed points of tangents to conics and their significance in coordinate geometry.
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Students studying conic sections, particularly those focusing on ellipses, as well as educators teaching coordinate geometry concepts related to tangents and chords of contact.

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Homework Statement



Show that the tangent to the ellipse ax^2+by^2=1 at the point (h,k) has equation ahx+bky=1

Hence, deduce that the chord of contact of tangents from the point (m,n) to the ellipse ax^2+by^2=1 has equation amx+bny=1

Homework Equations





The Attempt at a Solution



I managed to prove the first part but am having problem with the second part.

Of course, it can be done by evaluating the gradient of tangent of the ellipse and use the straight line formulas to prove that.

But i am not sure how to DEDUCE that from what i got from the first part.
 
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Any line through (m,n) is of the form y- n= a(x- m) where "a" is the slope of the line. On the other hand, any tangent to the ellipse at (h, k) has equation ahx+ bky= 1 which we can rewrite as bky= -ahx+ 1 or y= -(ah/bk)x+ 1/bk which has slope -(ah/bk). A line that both passes through (m,n) and is tangent to the ellipse at (h, k) must be of the form y- n= -(ah/bk)(x- m) or, multiplying through by bk, bky- bkn= -ahx+ ahm or ahx+ bky= bkn+ ahm.

Now use the fact that (h, k) is a point on the ellipse: [itex]ah^2+ bk^2= 1[/itex]
 
HallsofIvy said:
Any line through (m,n) is of the form y- n= a(x- m) where "a" is the slope of the line. On the other hand, any tangent to the ellipse at (h, k) has equation ahx+ bky= 1 which we can rewrite as bky= -ahx+ 1 or y= -(ah/bk)x+ 1/bk which has slope -(ah/bk). A line that both passes through (m,n) and is tangent to the ellipse at (h, k) must be of the form y- n= -(ah/bk)(x- m) or, multiplying through by bk, bky- bkn= -ahx+ ahm or ahx+ bky= bkn+ ahm.

Now use the fact that (h, k) is a point on the ellipse: [itex]ah^2+ bk^2= 1[/itex]

thanks, this is the continuatino of the question.

Show that for all values of t, the chord of contact of tangents from the point (2t, 1-t) to the ellipse ax^2+by^2=1 passes through a fixed point and determine the point of this coordinates.

The cartesian equation of (2t,1-t) is y=-1/2 x+1

So any point from this line to the ellipse will pass through a fixed point say (p,q)

Any further hints on this?
 

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