Find the angle at which the curves y=x² and x=tfrac53 cos

Find the angle at which the curves $y=x^{2}$ and $x=\tfrac53\cos t,\ y=\tfrac54\sin t$ intersect.

Step 1: Eliminate the parameter

\[ \left(\frac{x}{5/3}\right)^{2}+\left(\frac{y}{5/4}\right)^{2}=1 \;\Longrightarrow\; 9x^{2}+16y^{2}=25. \tag{1} \]

Step 2: Intersection points

Substituting $y=x^2$: $16x^4+9x^2-25=0$; with $u=x^2$, $(16u+25)(u-1)=0$, so the only real solution is $u=1$, giving $x=\pm1,\ y=1$. By symmetry, work at $P_1=(1,1)$.

Step 3: Slopes at $P_1$

Parabola: $k_1=y'(1)=2$. Ellipse, from (1): $18x+32yy'=0 \Rightarrow y'=-\dfrac{9x}{16y} \Rightarrow k_2=-\dfrac{9}{16}$.

Step 4: Angle

\[ \tan\theta = \left|\frac{2-\!\left(-\dfrac9{16}\right)}{1+2\!\left(-\dfrac9{16}\right)}\right| = \left|\frac{41/16}{-1/8}\right| = \frac{41}{2}. \tag{2} \]

Step 5: Verification

\[\text{At } P_2=(-1,1)\text{: } k_1=-2,\ k_2=\tfrac9{16},\ \tan\theta=\left|\tfrac{-2-9/16}{1-9/8}\right|=\tfrac{41}{2}\ \checkmark\]