Find angle between these two curves at point of intersection : $$K_1: x^2y^2 + y^4 = 1$$ and $$K_2 : x^2 + y^2 = 4 $$
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$\endgroup$ 34 Answers
$\begingroup$To solve the system, write $K_1$ as $y^2(x^2+y^2)=1$, then use $K_2$ to get $4y^2=1$.
$\endgroup$ $\begingroup$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*}
$\endgroup$ $\begingroup$$$ \theta = \arccos \frac{2}{\sqrt{19}} $$
HINT: Find the point of intersection and then find the tangent lines to each of the curves at that point. The angle between the lines is the angle between the curves at the point.
$\endgroup$ $\begingroup$Problem
Given the equations $$ \color{blue}{x^{2} + y^{2} = 4} \tag{1} $$ and $$ x^{2}y^{2} + y^{4} = 1 \tag{2} $$ find the angles between the curves at the points of intersection.
Symmetry analysis
Each term in both equations has a definite parity; this invites investigation.
The $x^{2}$ terms are invariant under the transformation $x\to-x$ because $$ (-x)^{2} = x^{2} $$ A similar argument holds for both $y^{2}$ and $y^{4}$.
From this we see invariance in form when both figures are reflected through each of the $x-$axis and the $y-$axis. Therefore, both figures are invariant under reflection through the origin. These ideas are demonstrated in the sequence of images below.
Conclusion: we only need to solve the problem at one intersection point.
Find intersections
Rewrite $(2)$: $$ y^{2} \left( x^{2} + y^{2} \right) = 1 \tag{2a} $$ and inserting $\color{blue}{(1)}$ we have $$ y^{2}\left( 4 \right) = 1 $$ or $$ y = \frac{1}{2} \tag{3} $$ Insert $(3)$ into $(1)$ to find $$ x = \frac{\sqrt{15}}{2}. $$
by the symmetry analysis, the locus of intersecting points is $$ \left\{ \frac{1}{2} \left( \sqrt{15}, 1 \right ), \frac{1}{2} \left( \sqrt{15}, -1 \right ), \frac{1}{2} \left( -\sqrt{15}, 1 \right ), \frac{1}{2} \left( -\sqrt{15}, -1 \right ) \right\} $$ We will focus on the point $p = \frac{1}{2} \left( \sqrt{15}, 1 \right )$.
Compute angles
Here are the slopes at the intersection point:
The slope calculation is detailed in the answer by @Manuel Guillen. Now to relate the slopes to angles.
Given the line $$ y = mx + b $$ the slope is related to the angle by $$ m = \tan \theta $$
The figure below shows the intersection angle as an arc between the tangent lines. The arc goes from $$ -\tan ^{-1}\left(\frac{\sqrt{15}}{17}\right) \le \theta \le \pi -\tan ^{-1}\left(\sqrt{15}\right) $$
The difference $\Delta$, is the angle of intersection:
$\endgroup$$$ \Delta = \pi +\tan ^{-1}\left(\frac{\sqrt{15}}{17}\right)-\tan ^{-1}\left(\sqrt{15}\right) = \pi -\cot ^{-1}\left(\frac{2}{\sqrt{15}}\right) \approx 117^{\circ} $$
