# 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?

Vikas TU
14149 Points
6 years ago
Given x3+y3=a3.

The derivative is,
• dy/dx=−x2 / y                                                                                    (1)

Therefore, slope of tangent at (x1,y1) is
• −x12 / y12                                                                                             (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) = −x12 / y12                                                                        (4.1)

{(y23 −y13) / (x23−x13) }× {(x12+ x1x2 +x22) / (y12+ y1y2 +y22)}= −x12 / y12         (4.2)

-{(x12+ x1x2 +x22) / (y12+ y1y2 +y22)}= −x12 / y12                                      (4.3)

x12 y12 +x1x2y12 +x22 y12 = x12 y12 +x12y1y2+x12 y22                                           (4.4)

x1x2y12 +x22 y12  = x12y1y2+x12 y22                                                                        (4.5)

x12 y22  - x22 y12  = x1x2y12  -  x12y1y2                                                                    (4.6)

(x1y2−x2y1)(x1y2+x2y1) = x1y1(x2y1−x1y2)                                                          (4.7)

x1y2+x2y1=−x1y1                                                                                                     (4.8)

(x2 / x1)+(y2 / y1) = −1                                                                                              (4.9)