x2+8y=0
y=−81x2
y′=−41x
y−y1=y′(x−x1)
y=−41x1(x−x1)−81x12
y=−81x1x+41x12
x+y−2=0=>y=−x+2
−41x1=−181x12=2 The system has the solution x1=4.
y1=−81(4)2=−2 The line x+y−2=0 is a tangent to the parabola x2+8y=0 at the point (4,−2).
The position vector of the point of contact is ⟨4,−2⟩.
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