Page 408 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 31: Joachimsthal's Notations

Tangent Pair Notes

If two points P(x , y ) and P(x , y ) are such that the line joining them is tangent to a conic s = 0,
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.
then as in (13), s = s s . Now, fixing P(x , y ) outside the conic and making P(x , y ) an
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arbitrary point on the tangent from P(x , y ), we can remove the second index:
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s = s s. ...(16)
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The latter is a quadratic equation which may be factorized into the product of two linear equations
each representing a tangent to the conic through P(x , y ).
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Example:
Let s = x + 4y - 25, so that s = 0 is an ellipse x + 4y = 25. What are the tangents from P(0, 0) to the
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ellipse? Let’s see that there are none. First,
s = -25 and s = -25.
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So that (16) becomes
s = -25, orx + 4y = 0.
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Obviously the equation has no real roots (besides x = y = 0), nor linear factors. We conclude that
there are no tangents from (0, 0) to the ellipse. Naturally. But let’s now take a different point, say
P(5, 5/2). In this case,
s = 25 and s = 5x + 10y - 25.
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(16) then becomes
(5x + 10y - 25) = 25·(x + 4y - 25).
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First, let’s simplify this to
(x + 2y - 5) = x + 4y - 25.
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Second, let’s multiply out and simplify by collecting the like terms:
2xy - 5x - 10y + 25 = 0,
which is factorized into
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(x - 5) (2y - 5) = 0.
Conclusion: here are two tangents from (5, 5/2) to the ellipse: x = 5 and y = 5/2.
Poles and Polars With Respect To a Conic
Let P(x , y ) be a point outside a conic s = 0 and P(x , y ) and P(x , y ) be the points where the
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tangents from P(x , y ) meet the conic.
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Then the tangents have the equations (15)
s = 0 and s = 0 ...(17)
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and also meet at P(x , y ):
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s = 0 and s = 0. ...(18)
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Because of the symmetry of the notations, we have
s = 0 and s = 0, ...(19)
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