At which point of the parabola y = x²-2x+5 must
At which point of the parabola $y = x^{2}-2x+5$ must the tangent be drawn so that it is perpendicular to the line $y = x$?
Explanation of Method
Perpendicularity fixes the slope at −1.
Step 1: The required slope
The line $y = x$ has slope $1$; a line perpendicular to it has slope
\[ k = -\frac11 = -1. \]
Step 2: The point of tangency
\[ y' = 2x-2 = -1 \;\Longrightarrow\; x_{0} = \frac12, \qquad y_{0} = \frac14-1+5 = \frac{17}{4}. \]
Step 3: The tangent
\[ y-\frac{17}{4} = -\left(x-\frac12\right) \;\Longrightarrow\; y = -x+\frac{19}{4}. \]
Step 4: Verification
\[\left(\tfrac12\right)^{2}-2\cdot\tfrac12+5 = \tfrac14-1+5 = \tfrac{17}{4} \;\;\checkmark\]
\[y'\!\left(\tfrac12\right) = -1, \qquad 1\cdot(-1) = -1 \;\;\checkmark\]
\[\text{The vertex is at } x = 1;\ \text{the point found lies half a unit to its left} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\text{point } \left(\frac12,\;\frac{17}{4}\right), \qquad \text{tangent } y = -x+\frac{19}{4}\,}\]
Remark
Remark. Since the tangent is perpendicular to $y = x$, the normal at that point is parallel to $y = x$ — the two phrasings describe the same point. Note also that the tangent found, $y = -x+\tfrac{19}{4}$, meets $y = x$ at $x = \tfrac{19}{8}$, well away from the parabola.