On the curve y = x²-1 find the point at which 1°
On the curve $y = x^{2}-1$ find the point at which 1° the normal is parallel to the line $y = x+1$; 2° the normal is perpendicular to the line $y = -x$.
Explanation of Method
Two conditions that turn out to be the same condition.
Step 1: The slope of the normal
With $y' = 2x$, the normal at $\bigl(x_{0},x_{0}^{2}-1\bigr)$ has slope
\[ k_{n} = -\frac{1}{2x_{0}} \qquad \bigl(x_{0}\neq0\bigr). \tag{1} \]
Step 2: 1° — parallel to $y = x+1$
Parallel means $k_{n} = 1$: \[-\frac{1}{2x_{0}} = 1 \;\Longrightarrow\; x_{0} = -\frac12, \qquad y_{0} = \frac14-1 = -\frac34,\]
\[ \text{normal: } y+\frac34 = x+\frac12 \;\Longrightarrow\; y = x-\frac14. \]
Step 3: 2° — perpendicular to $y = -x$
Perpendicular to a line of slope $-1$ means $k_{n}\cdot(-1) = -1$, i.e. $k_{n} = 1$ — the very same requirement. Hence the same point and the same normal.
Step 4: Verification
\[\left(-\tfrac12\right)^{2}-1 = -\tfrac34 \;\;\checkmark; \qquad y'\!\left(-\tfrac12\right) = -1 \;\Longrightarrow\; k_{n} = 1 \;\;\checkmark\]
\[1 = 1 \ \text{(parallel to } y = x+1); \qquad 1\cdot(-1) = -1 \ \text{(perpendicular to } y = -x) \;\;\checkmark\]
\[\text{The tangent there is } y = -x-\frac54,\ \text{perpendicular to the normal} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\text{both conditions give } \left(-\frac12,\;-\frac34\right), \qquad \text{normal } y = x-\frac14\,}\]
Remark
Remark. The lines $y = x+1$ and $y = -x$ are themselves perpendicular, so "parallel to the first" and "perpendicular to the second" are two ways of saying the same thing about a third line. The exercise is really a check on the bookkeeping of slopes.