Q. \( x^4 + y^4 – 2a^2\left(x^2 + y^2\right) + a^4 = 0 \).

Q: How can the polynomial be factored?

It factors as two quadratic factors: x^4+y^4-2a^2(x^2+y^2)+a^4=(x^2+y^2-a^2-\sqrt2\,xy)(x^2+y^2-a^2+\sqrt2\,xy)..

Q: Where does the curve meet the coordinate axes?

Set y=0 or x=0: \left(x^2-a^2\right)^2=0. So the curve touches axes at \left(\pm a,0\right) and \left(0,\pm a\right) (double/contact points)..

Q: How to solve it in polar coordinates?

With x=r\cos\theta, y=r\sin\theta you get r^4(\cos^4\theta+\sin^4\theta)-2a^2r^2+a^4=0. Solve the quadratic in r^2: r^2=\dfrac{a^2\big(1\pm\sqrt{1-(\cos^4\theta+\sin^4\theta)}\big)}{\cos^4\theta+\sin^4\theta}.

Q: How do I remove the xy term (rotation angle)?

For Ax^2+Bxy+Cy^2+\cdots with A=C=1,B=\pm\sqrt2, the rotation angle satisfies \tan 2\phi=B/(A-C) which is infinite, so \phi=\pi/4. The ellipses are rotated by 45^\circ..

Q: Are there special parameter cases (e.g., a=0)?

Are there special parameter cases (e.g., a=0)?

Q: How to find explicit Cartesian equations of each ellipse?.

Each factor is x^2+y^2\pm\sqrt2\,xy=a^2. Diagonalize by rotating 45^\circ: set u=(x+y)/\sqrt2,\ v=(x-y)/\sqrt2. Then each becomes a standard ellipse in u,v (compute coefficients to identify semi-axes).

Q: Is this a lemniscate or related to known named curves?

No; it's not the Bernoulli lemniscate. It is simply the union of two rotated ellipses (degree-4 overall but reducible into two degree-2 factors)..

Answer

\[
x^4+y^4-2a^2(x^2+y^2)+a^4=(x^2+y^2)^2-2x^2y^2-2a^2(x^2+y^2)+a^4=(x^2+y^2-a^2)^2-2x^2y^2
\]
\[
=(x^2+y^2-a^2-\sqrt{2}\,xy)(x^2+y^2-a^2+\sqrt{2}\,xy)=0
\]
Thus the solutions are the points on the two conics
\[
x^2+y^2-a^2\pm\sqrt{2}\,xy=0.
\]

Detailed Explanation

We will algebraically rearrange the given equation into a recognizable geometric form by completing the square.

\[x^4 + y^4 – 2a^2(x^2 + y^2) + a^4 = 0\]

First, distribute the \( -2a^2 \) term to expand the expression, resulting in \( x^4 – 2a^2 x^2 + y^4 – 2a^2 y^2 + a^4 = 0 \).

Next, we complete the square for the \( x \) terms. Group \( x^4 – 2a^2 x^2 \) and pair it with the existing \( a^4 \) to form a perfect square trinomial. This groups as \( (x^4 – 2a^2 x^2 + a^4) + y^4 – 2a^2 y^2 = 0 \).

This simplifies nicely to \( (x^2 – a^2)^2 + y^4 – 2a^2 y^2 = 0 \).

Now, we complete the square for the remaining \( y \) terms. We have \( y^4 – 2a^2 y^2 \) and need to add \( a^4 \) to make it a perfect square. To maintain balance, we must add \( a^4 \) to both sides of the equation, giving \( (x^2 – a^2)^2 + (y^4 – 2a^2 y^2 + a^4) = a^4 \).

This simplifies to the final equivalent form of \( (x^2 – a^2)^2 + (y^2 – a^2)^2 = a^4 \).

In the \( x^2, y^2 \) coordinate plane, this represents a perfect circle centered at \( (a^2, a^2) \) with a radius of \( a^2 \). In the standard \( x, y \) Cartesian plane, this maps to a bounded, symmetrical algebraic curve.

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Algebra FAQs

How can the polynomial be factored?

It factors as two quadratic factors: \(x^4+y^4-2a^2(x^2+y^2)+a^4=(x^2+y^2-a^2-\sqrt2\,xy)(x^2+y^2-a^2+\sqrt2\,xy)\)..

What geometric curve does the equation represent?

It is the union of two conics (each quadratic factor = 0). Since each has discriminant \(B^2 - 4AC = 2 - 4 = -2 < 0\), both are ellipses (rotated, centered at the origin).

Where does the curve meet the coordinate axes?

Set \(y=0\) or \(x=0\): \(\left(x^2-a^2\right)^2=0\). So the curve touches axes at \(\left(\pm a,0\right)\) and \(\left(0,\pm a\right)\) (double/contact points)..

How to solve it in polar coordinates?

With \(x=r\cos\theta, y=r\sin\theta\) you get \(r^4(\cos^4\theta+\sin^4\theta)-2a^2r^2+a^4=0\). Solve the quadratic in \(r^2\): \(r^2=\dfrac{a^2\big(1\pm\sqrt{1-(\cos^4\theta+\sin^4\theta)}\big)}{\cos^4\theta+\sin^4\theta}\).

How do I remove the \(xy\) term (rotation angle)?

For \(Ax^2+Bxy+Cy^2+\cdots\) with \(A=C=1,B=\pm\sqrt2\), the rotation angle satisfies \(\tan 2\phi=B/(A-C)\) which is infinite, so \(\phi=\pi/4\). The ellipses are rotated by \(45^\circ\)..

Are there special parameter cases (e.g., \(a=0\))?

How to find explicit Cartesian equations of each ellipse?.

Each factor is \(x^2+y^2\pm\sqrt2\,xy=a^2\). Diagonalize by rotating \(45^\circ\): set \(u=(x+y)/\sqrt2,\ v=(x-y)/\sqrt2\). Then each becomes a standard ellipse in \(u,v\) (compute coefficients to identify semi-axes).

Is this a lemniscate or related to known named curves?