KGS math club/solution 10 1: Difference between revisions
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The coefficient of the tangent can be found from implicit derivative formula: | The coefficient of the tangent can be found from implicit derivative formula: | ||
So we want to find a pair (x, y) such that <math> x^2 + y^2 + x y - 1 = 0 </math> and <math> (y - 2) / x = - (2x + y) / (2y + x) </math> | <math>dy / dx = - (dz / dx) / (dz / dy) = - (2x + y) / (2y + x) </math> | ||
where <math> z = x^2 + y^2 + x y </math> | |||
So we want to find a pair (x, y) such that | |||
<math> x^2 + y^2 + x y - 1 = 0 </math> | |||
and | |||
<math> (y - 2) / x = - (2x + y) / (2y + x) </math> | |||
<math> <=> 2 y^2 + (x - 4 + x) y - 2 x + 2 x^2 = 0 </math> | |||
<math> <=> y^2 + (x - 2) y - x + x^2 = 0 </math> | |||
We notice by magic that (x, y) = (1, 0) and (x, y) = (-1, 1) are the two solutions to the equation. | We notice by magic that (x, y) = (1, 0) and (x, y) = (-1, 1) are the two solutions to the equation. | ||
Verification: | Verification: | ||
at (-1, 1), the <math> dy / dx = - (-2 + 1) / (2 - 1) = 1 </math>, so the tangent goes from (-1, 1) to (0, 2) | at (-1, 1), the <math> dy / dx = - (-2 + 1) / (2 - 1) = 1 </math>, so the tangent goes from (-1, 1) to (0, 2) | ||
at (1, 0), the <math> dy / dx = -(2 + 0) / (0 + 1) = -2 </math>, so the tangent goes from (1, 0) to (0, 2) | at (1, 0), the <math> dy / dx = -(2 + 0) / (0 + 1) = -2 </math>, so the tangent goes from (1, 0) to (0, 2) | ||
'''This solution is incomplete, please fill in!''' | '''This solution is incomplete, please fill in!''' | ||
Revision as of 14:02, 11 August 2010
The coefficient of the tangent can be found from implicit derivative formula:
where 
So we want to find a pair (x, y) such that
and
We notice by magic that (x, y) = (1, 0) and (x, y) = (-1, 1) are the two solutions to the equation.
Verification:
at (-1, 1), the 
, so the tangent goes from (-1, 1) to (0, 2)
at (1, 0), the 
, so the tangent goes from (1, 0) to (0, 2)
This solution is incomplete, please fill in!
