Finding the locus of $b^2-2x^2=2xy+y^2$

Question

I was asked to determine the locus of the equation $$b^2-2x^2=2xy+y^2$$

This is my work:

Add $x^2$ to both sides: $$\begin{align} b^2-x^2 &=2xy+y^2+x^2\\ b^2-x^2 &=\left(x+y\right)^2 \end{align}$$

I see that this is similar to the equation of a circle. How can I find the locus of this expression?

Answer 1

The matrix associated to the conic is $$\begin{pmatrix}-2 & -1 & 0\\-1 & -1 & 0\\0 & 0 & b^2 \end{pmatrix}\implies\Bigg|\begin{pmatrix}-2 & -1\\-1 & -1 \end{pmatrix}\Bigg|=2-1=1>0$$ and this shows that the conic is an ellipse.

Answer 2

HINT.-A way to study the curve is to consider it as a union of two graphs of functions (similar to how you would see two functions $y=\pm\sqrt{r^2-x^2}$ for a circle) so you get two explicits functions: $$y=-x+\sqrt{b^2-x^2}\\y=-x-\sqrt{b^2-x^2}$$ whose domains are $[-b,b]$.

This way you have an ellipse for each value of $b$ whose axis are (approximately) in the lines $$y=0.625x\\y-1.6x$$

For example, for $b=2$ you have (approximately) the semiaxis $1.2358$ and $3.2358$.

Answer 3

The equation is clearly of a conic section. However,since the coefficient of $xy$ is non-zero, the conic section is tilted by some angle $\theta$. The value of $\theta$ can be determined as: $$\theta = \dfrac{1}2 \arctan (\dfrac{C}{A-B})$$ , where $A,\ C $ and $B$ are the coefficients of $x^2$, $y^2$ and $xy$ respectively in the equation of the conic section. Substitution yields $\theta = 0.5535\dots$ (in radians). Now rotate the conic section by $-\theta$ to get its standard equation with vertical or horizontal orientation. $$ x = x^\prime \cos \theta - y^\prime \sin \theta $$ $$y = x^\prime \sin \theta + y^\prime \cos \theta$$

After substitution and (ugly) computation, you will get the following equation:

$$2.62 (x')^2 + 0.38(y')^2 = b^2$$ This means the original equation is an equation of an ellipse with major axis of length $a^\prime= b\sqrt{1/0.38}$ tilted by $.5535\dots$ radians (about $31$ degrees) and minor axis of lengh $b^\prime = b\sqrt{1/2.62}$

But,of course, there are easier tests you can check with what conic section formula it is.

Answer 4

With variable changes $$x= \sqrt{ \frac{5+\sqrt5}2}u + \sqrt{ \frac{5-\sqrt5}2}v,\>\>\>\>\> y= \sqrt{ \frac{5-\sqrt5}2}u - \sqrt{ \frac{5+\sqrt5}2}v $$ the curve equation $b^2-2x^2=2xy+y^2$ can be recast as

$$ \frac{u^2}{\left(\frac{\sqrt5-1}2b\right)^2} +\frac{v^2}{\left(\frac{\sqrt5+1}2b\right)^2} =1 $$ which reveals an ellipse of axis lengths $\frac{\sqrt5-1}2b$ and $\frac{\sqrt5+1}2b$.

Answer 5

The equation $$Ax^2 + Bxy+Cy^2 +Dx + Ey + F = 0$$ is:

• an ellipse if $B^2 - 4AC < 0$ (could be circle or a point if $A = C$, $F > 0$ and $B = 0$)

• a parabola if $B^2 -4AC = 0$

• a hyperbola if $B^2 - 4AC > 0$

In this case, $B^2-4AC = -4 <0$. Hence it is an ellipse.