Find the angle at which the curves x=t, y=t² and x=5/3

Find the angle at which the curves $x=t,\ y=t^{2}$ and $x=\dfrac{5}{3}\cos t,\ y=\dfrac{5}{4}\sin t$ intersect.

Step 1: Identify the two curves

The first is simply the parabola $y = x^{2}$. For the second, eliminate $t$ by the Pythagorean identity: \[\left(\frac{3x}{5}\right)^{2} + \left(\frac{4y}{5}\right)^{2} = \cos^{2}t + \sin^{2}t = 1 \;\Longrightarrow\; 9x^{2}+16y^{2} = 25,\]

an ellipse with semi-axes $\tfrac53$ and $\tfrac54$.

Step 2: Find the intersection points

Substituting $y = x^{2}$ into the ellipse gives a quadratic in $u = x^{2}$: \[9x^{2}+16x^{4} = 25 \;\Longrightarrow\; 16u^{2}+9u-25 = 0 \;\Longrightarrow\; u = \frac{-9\pm\sqrt{81+1600}}{32} = \frac{-9\pm 41}{32}.\]

The negative root is rejected since $u = x^{2} \geq 0$; the positive one is $u = 1$, so

\[ x = \pm 1, \qquad y = 1: \qquad P_{1}(1,1), \quad P_{2}(-1,1). \]

Step 3: Slopes at $P_{1}(1,1)$

Parabola: $y' = 2x$, so $k_{1} = 2$.

Ellipse: differentiating $9x^{2}+16y^{2}=25$ implicitly,

\[ 18x + 32y\,y' = 0 \;\Longrightarrow\; y' = -\frac{9x}{16y}, \qquad k_{2} = -\frac{9}{16}. \]

Step 4: Apply the angle formula

\[ \tan\alpha = \left|\frac{k_{1}-k_{2}}{1+k_{1}k_{2}}\right| = \left|\frac{2+\tfrac{9}{16}}{1-\tfrac{9}{8}}\right| = \left|\frac{\tfrac{41}{16}}{-\tfrac18}\right| = \frac{41}{2}. \]

\[ \alpha = \arctan\frac{41}{2} = \arctan 20.5 \approx 87.207^{\circ}. \]

Step 5: The second point

At $P_{2}(-1,1)$ both slopes flip sign: $k_{1} = -2$ and $k_{2} = \tfrac{9}{16}$, so

\[ \tan\alpha = \left|\frac{-2-\tfrac{9}{16}}{1-\tfrac98}\right| = \left|\frac{-\tfrac{41}{16}}{-\tfrac18}\right| = \frac{41}{2}, \]

the same angle — as the symmetry of both curves about the $y$-axis requires.

Step 6: Verification

\[P_{1} \text{ on the parabola: } 1 = 1^{2} \;\checkmark; \quad \text{on the ellipse: } 9+16 = 25 \;\checkmark\]

\[87.207^{\circ} = 87^{\circ} + 0.207\times 60' \approx 87^{\circ}12' \;\;\checkmark\]