# Find the point of intersection of the tangents to the curve `y^2 -3xy + x^3 = 3` at the points where `x=-1.`

## Expert Answers

Hello!

The steps are: 1) find the tangent points, 2) find the equations of the tangents, 3) find the intersection of the tangents.

1. It is known that `x=-1.` Substitute it to the curve equation and obtain an equation for `y:`

`y^2+3y-1=3,` or `y^2+3y-4=0.`

The solutions are `y_1=1` and `y_2=-4.` So the points of tangent are `T_1(-1,1)` and `T_2(-1,-4).`

2. Differentiate the equation `d/(dx):`

`2yy' - 3y - 3xy' + 3x^2 = 0,` or `y'(2y - 3x) = 3y - 3x^2,` or

`y'=(3(y-x^2))/(2y-3x).`

For `T_1` we obtain that `y'=0.` Thus the tangent is a horizontal line with the equation `y=1.`

For `T_2` we obtain `y'=(3(-4 - 1))/(2*(-4) - 3*(-1)) = -15/(-5) = 3.` Thus the equation is `y=3(x+1)-4 = 3x-1.`

3. The point of intersection has `y=1` and `x` from the equation `1=3x-1,` i.e. `x=2/3.` So the answer is the point `(2/3, 1).`

The graph is at the link.

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