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$$5x^2+8xy+5y^2=1$$

$$1\left(\frac{x-y}{\sqrt{2}}\right)^2+9\left(\frac{x+y}{\sqrt{2}}\right)^2=1$$

I know that these two forms are equal, showing that the equation is an ellipse.

I do know what happens when the ellipse is in the form of $ax^2 + by^2 = 1$ but not sure what happens in the combination of x-y cases.

Can anyone help me with this? (e.g., what would be two axis?)

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3 Answers 3

up vote 3 down vote accepted

In the familiar case $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the axes of the ellipse are $x=0$ and $y=0$. In this case, they are $x-y=0$ and $x+y=0$.

What we have here is a "standard" ellipse, rotated through $45^\circ$.

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One way to understand this is that u make two graphs , each of which related to one part $((x-y)^2/c)$ and $((x-y)^2)/d$ and then add both parts and add both corresponding graphs.c and d are constants.

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Using this, if we rotate the axes $\theta $ as the rotation of axes does not change the nature of the curve,

$x=X\cos\theta-Y\sin\theta$, $y=X\sin\theta+Y\cos\theta$,

$5(X\cos\theta-Y\sin\theta)^2+5(X\sin\theta+Y\cos\theta)^2+8(X\cos\theta-Y\sin\theta)(X\sin\theta+Y\cos\theta)=1$

$=>X^2(5+8\sin\theta\cos\theta)+Y^2(5-8\sin\theta\cos\theta)+8XY(\cos^2\theta-\sin^2\theta)=1$

$=>X^2(5+4\sin2\theta)+Y^2(5-4\sin2\theta)+8XY(\cos2\theta)=1$

If we make $\cos2\theta=0=>\sin2\theta=±1$

If $\sin2\theta=1$ i.e, $2\theta=2n\pi+\frac{\pi}{2}=>9X^2+Y^2=1$

If $\sin2\theta=-1$ i.e, $2\theta=2n\pi-\frac{\pi}{2}=>X^2+9Y^2=1$


Alternatively, according to 7.2 of this,

Here a=b=5, h=4, f=g=0

So,

$C=\begin{pmatrix} a & h \\ h & b \end{pmatrix}=5^2-4^2>0$

$Δ= \begin{pmatrix} a & h & g \\ h & b & f \\g & f & c \end{pmatrix}$ $=\begin{pmatrix} 5 & 4 & 0 \\ 4 & 5 & 0 \\0 & 0 & -1 \end{pmatrix}$ $=-(5^2-4^2)=-9<0$

and $aΔ=5(-9)<0$

So, the given curve represents an ellipse.

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