KGS math club/solution 10 1

The coefficient of the tangent can be found from implicit derivative formula:

$dy / dx = - (dz / dx) / (dz / dy) = - (2x + y) / (2y + x)$

where $z = x^2 + y^2 + x y$

So we want to find a pair (x, y) such that

$x^2 + y^2 + x y - 1 = 0$

and

$(y - 2) / x = - (2x + y) / (2y + x)$

$<=> 2 y^2 + (x - 4 + x) y - 2 x + 2 x^2 = 0$

$<=> y^2 + (x - 2) y - x + x^2 = 0$

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 $dy / dx = - (-2 + 1) / (2 - 1) = 1$ , so the tangent goes from (-1, 1) to (0, 2)

at (1, 0), the $dy / dx = -(2 + 0) / (0 + 1) = -2$ , so the tangent goes from (1, 0) to (0, 2)

This solution is incomplete, please fill in!

Page