If a tangent at point (x1,y1) to a curve x^3+y^3=a^3 meets the curve again at point (x2,y2), how does one prove that x2/x1 + y2/y1 = - 1?

Question

Vikas TU · Answer

Given x 3 +y 3 =a 3 . The derivative is, dy/dx=−x 2 / y 2 (1) Therefore, slope of tangent at (x1,y1) is −x1 2 / y1 2 (2) The tangent passes through (x2,y2), therefore, slope of the tangent is also given by (y2−y1) / (x2−x1) (3) Comparing the two slope equations we get, (y2−y1) / (x2−x1) = −x1 2 / y1 2 (4.1) {(y2 3 −y1 3 ) / (x2 3 −x1 3 ) }× {(x1 2 + x1x2 +x2 2 ) / (y1 2 + y1y2 +y2 2 )}= −x1 2 / y1 2 (4.2) -{(x1 2 + x1x2 +x2 2 ) / (y1 2 + y1y2 +y2 2 )}= −x1 2 / y1 2 (4.3) x1 2 y1 2 +x1x2y1 2 +x2 2 y1 2 = x1 2 y1 2 +x1 2 y1y2+x1 2 y2 2 (4.4) x1x2y1 2 +x2 2 y1 2 = x1 2 y1y2+x1 2 y2 2 (4.5) x1 2 y2 2 - x2 2 y1 2 = x1x2y1 2 - x1 2 y1y2 (4.6) (x1y2−x2y1)(x1y2+x2y1) = x1y1(x2y1−x1y2) (4.7) x1y2+x2y1=−x1y1 (4.8) (x2 / x1)+(y2 / y1) = −1 (4.9)

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If a tangent at point (x1,y1) to a curve x^3+y^3=a^3 meets the curve again at point (x2,y2), how does one prove that x2/x1 + y2/y1 = - 1?

If a tangent at point (x1,y1) to a curve x^3+y^3=a^3 meets the curve again at point (x2,y2), how does one prove that x2/x1 + y2/y1 = - 1?

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If a tangent at point (x1,y1) to a curve x^3+y^3=a^3 meets the curve again at point (x2,y2), how does one prove that x2/x1 + y2/y1 = - 1?

If a tangent at point (x1,y1) to a curve x^3+y^3=a^3 meets the curve again at point (x2,y2), how does one prove that x2/x1 + y2/y1 = - 1?

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