<s>[INST] Find Angle Between Two Curves at Point of Intersection Find angle between these two curves at point of intersection : 
$$K_1: x^2y^2 + y^4 = 1$$ and 
$$K_2 : x^2 + y^2 = 4 $$ 
Thanks!
 [/INST] To find the points of intersection, we see that the equation of the curve $K_1$ can be written as:
\begin{align*}
x^2 y^2 + y^4 &= 1 \\
y^2 (x^2 + y^2) &= 1
\end{align*}
Since the points of intersection lies on both curve $K_1$ and $K_2$, it must satisfy both the above equation and the equation of $K_2$, meaning:
\begin{align*}
y^2 (x^2 + y^2) &= 1 \\
y^2 \cdot 4 &= 1 \\
y^2 &= \frac{1}{4}
\end{align*}
Plugging this back into the equation for $K_1$, we get:
\begin{align*}
x^2 + \frac{1}{4} &= 4 \\
x^2 &= \frac{15}{4}
\end{align*}
So we have four points of intersection, corresponding to the four combinations of:
$$ x = \pm \frac{\sqrt{15}}{2} \qquad y = \pm \frac{1}{2} $$
The angle of intersection of the curves is the same as the angle between the tangent line of $K_1$ and the tangent line of $K_2$ at the point of intersection.
Looking at the slope of the curve $K_1$, we take the equation of the curve and perform implicit differentiation:
\begin{align*}
x^2 y^2 + y^4 &= 1 \\
\frac{d}{dx} \left[ x^2 y^2 + y^4 \right] &= \frac{d}{dx} 1 \\
2x y^2 + 2 x^2 y \frac{dy}{dx} + 4y^3 \frac{dy}{dx} &= 0 \\
\frac{dy}{dx} &= -\frac{2xy^2}{2x^2 y + 4y^3} \\
&= -\frac{xy}{x^2 + 2y^2}
\end{align*}
Looking at the slope of the curve $K_2$, seeing as it is a circle, we can verify that implicit differentiation gives us:
$$ \frac{dy}{dx} = - \frac{x}{y} $$
Regardless of which of the four intersection points we choose, the slopes have the same sign, as seen by the formulas, so we have two possible pairs of values for the slopes:
$$ (m_1,m_2) = \pm \left(\frac{\sqrt{15}}{17} , \sqrt{15} \right)$$
Turning these slopes into angles of inclinations by taking the arctangent, we get:
$$ (\theta_1, \theta_2) = \pm \left( \arctan \frac{\sqrt{15}}{17} , \arctan \sqrt{15} \right)$$
The angle between the tangent lines is the difference between the inclination angles of the tangent lines, which results in:

$$ \theta = \arctan \sqrt{15} - \arctan \frac{\sqrt{15}}{17} $$

Another way to get an equivalent answer is to deal with the normal vectors of the tangent lines and find the angle between the vectors. Since we have seen that the intersection angle at all four points is the same, the angle we care about can be seen as the angle between the normal vectors:
$$ \vec{n}_1 = \langle \sqrt{15}, 1 \rangle \qquad \vec{n}_2 = \left\langle \frac{\sqrt{15}}{17}, 1 \right\rangle $$
which ends up resulting in:
\begin{align*}
\cos \theta &= \frac{\vec{n}_1 \cdot \vec{n}_2}{\lVert \vec{n}_1 \rVert \lVert \vec{n}_2 \rVert} \\
&= \frac{\frac{15}{17} + 1}{4 \cdot \frac{4 \sqrt{19}}{17}} \\
&= \frac{15 + 17}{16 \sqrt{19}} \\
&= \frac{2}{\sqrt{19}}
\end{align*}

$$ \theta = \arccos \frac{2}{\sqrt{19}} $$

</s>