Find the angle between the tangents drawn

Find the angle between the tangents drawn at the intersection point of the curves $y = \dfrac{1}{x}$ and $y = x^{2}$.

Step 1: Find the intersection point

Equating the two expressions for $y$: \[\frac{1}{x} = x^{2} \;\Longleftrightarrow\; x^{3} = 1 \;\Longleftrightarrow\; x = 1,\]

the only real solution. The corresponding ordinate is $y = 1$, so the curves meet at the single point

\[ P = (1,\,1). \]

Step 2: Find the slopes of the two tangents at $P$

First curve: $y = \dfrac{1}{x} = x^{-1}$, so $y' = -\dfrac{1}{x^{2}}$ and

\[ k_{1} = y'\big|_{x=1} = -1. \]

Second curve: $y = x^{2}$, so $y' = 2x$ and

\[ k_{2} = y'\big|_{x=1} = 2. \]

Step 3: Apply the angle formula

The angle $\alpha$ between two lines of slopes $k_{1}$ and $k_{2}$ satisfies

\[ \tan\alpha = \left|\frac{k_{2}-k_{1}}{1+k_{1}k_{2}}\right|. \]

Substituting $k_{1} = -1$ and $k_{2} = 2$: \[\tan\alpha = \left|\frac{2-(-1)}{1+(-1)(2)}\right| = \left|\frac{3}{1-2}\right| = \left|\frac{3}{-1}\right| = 3.\]

Step 4: Compute the angle

\[ \alpha = \arctan 3 \approx 71.565^{\circ}. \]

Step 5: Verification

\[k_{1} \neq k_{2} \Rightarrow \text{not parallel}; \qquad k_{1}k_{2} = -2 \neq -1 \Rightarrow \text{not perpendicular}\]

\[0.565^{\circ} \times 60 \approx 34' \;\Longrightarrow\; \alpha \approx 71^{\circ}34' \;\;\checkmark\]

The angle lies strictly between $0^{\circ}$ and $90^{\circ}$, as the two conditions above require.