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Can someone help me with a weird calculus question?
The line y=dx+e is a tangent line to the parabola y=ax^2+bx+c
Prove that: (b-d)^2=4a(c-e)
1 Answer
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- ?Lv 74 years agoFavorite Answer
y = ax^2 + bx + c = dx + e
ax^2 + (b – d)x + c – e = 0
The usual two solutions to a quadratic are in this case coincident,
because the tangent touches the parabola in a single point.
This means the discriminant is zero, and hence
(b - d)^2 - 4a(c - e) = 0
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