Difference between revisions of "KGS math club/solution 10 1"
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So we want to find a pair (x, y) such that | So we want to find a pair (x, y) such that | ||
| − | <math> x^2 + y^2 + x y | + | <math> x^2 + y^2 + x y = 1 </math> |
and | and | ||
| Line 15: | Line 15: | ||
<math> <=> 2 y^2 + (x - 4 + x) y - 2 x + 2 x^2 = 0 </math> | <math> <=> 2 y^2 + (x - 4 + x) y - 2 x + 2 x^2 = 0 </math> | ||
| − | <math> <=> y^2 + | + | <math> <=> y^2 + x y - 2 y - x + x^2 = 0 </math> |
| − | We | + | Substituting the first to the second, we get |
| + | |||
| + | <math> (y^2 + x y + x^2) - 2 y - x = 1 - 2 y - x = 0 </math> | ||
| + | |||
| + | <math> <=> y = (1 - x) / 2 </math> | ||
| + | |||
| + | Substituting this back to the first equation, we get | ||
| + | |||
| + | <math> x^2 + (1 - x)^2 / 4 + x (1 - x) / 2 - 1 = (1 + 1/4 - 1/2) x^2 + (-1/2 + 1/2) x + (1/4 - 1) = 3/4 x^2 - 3/4 = 0 </math> | ||
| + | |||
| + | <math> <=> x = +-1 </math> | ||
| + | |||
| + | We have thus found two solutions: <math>(x, y) = (1, 0)</math> and <math>(x, y) = (-1, 1)</math>. | ||
Verification: | Verification: | ||
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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) | ||
| − | |||
| − | |||
Latest revision as of 14:11, 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
Substituting the first to the second, we get
Substituting this back to the first equation, we get
We have thus found two solutions: 
and 
.
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)
