Prove that the curve given by $f(t)=\cos(t) u_1 + \sin(t)u_2$ is an ellipse. Let $u_1$, $u_2 \in \mathbb{R}^2$ be two linearly independent vectors. Prove that the curve given by $f(t)=\cos(t) u_1 + \sin(t)u_2$ is an ellipse.
My attempt
Let $u_1=(a_1,a_2)$, $u_2=(b_1,b_2)$, $x=a_1 \cos(t)+b_1 \sin(t)$, $y=a_2 \cos(t)+b_2 \sin(t)$
Then:
$x^2=a_1^2 \cos ^2(t) + a_1 b_1 \sin(2t) + b_1^2 \sin^2(t)$
$y^2=a_2^2 \cos ^2(t) + a_2 b_2 \sin(2t) + b_2^2 \sin^2(t)$
I'm trying to figure out what would be the two constants $\alpha, \beta$ such that $\dfrac{x^2}{\alpha^2}+ \dfrac{y^2}{\beta^2}=1$, but I'm not sure what to do next. I'm aware that the fact that $u_1$ and $u_2$ are linearly independent implies that $f(t) \neq 0$...
 A: First you should change from parametric to implicit form$$\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}a_1&b_1\\a_2&b_2\end{bmatrix}\begin{bmatrix}\cos t\\\sin t\end{bmatrix}.$$
Let $D=a_1b_2-b_1a_2$. Then$$\frac{1}{D}\begin{bmatrix}b_2&-b1\\-a_2&a_1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}\cos t\\\sin t\end{bmatrix},$$
$$\begin{bmatrix}\cos t&\sin t\end{bmatrix}\begin{bmatrix}\cos t\\\sin t\end{bmatrix}=[1]$$ $$\frac{1}{D^2}\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}b_2&-a_2\\-b_1&a_1\end{bmatrix}\begin{bmatrix}b_2&-b_1\\-a_2&a_1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=[1],$$ $$\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}b_2^2+a_2^2&-b_1b_2-a_1a_2\\-b_1b_2-a_1a_2&b_1^2+a_1^2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}D^2\end{bmatrix}.$$
Now that you have the conic in implicit form, you can rotate it into standard form and thence categorize it as an ellipse, hyperbola or parabola.
A: The answer follows from the geometric interpretation of the singular value decomposition (SVD). See the wikipedia page, in particular the first picture.
Let $A = [u_1\; u_2]$. Your curve is the image of the unit circle under the linear map $A$:
$$
    A \begin{bmatrix} \cos t \\ \sin t \end{bmatrix} = 
 \begin{bmatrix}
a_1 & b_1 \\
a_2 & b_2
\end{bmatrix}
 \begin{bmatrix} \cos t \\ \sin t \end{bmatrix}.
$$
Let $A = U\Sigma V^T$ be the SVD. The matrices $U$ and $V$ are orthogonal; they represent rotations or reflections. The matrix $\Sigma$ is a diagonal matrix with non-negative elements on the diagonal.
First, $V^T$ performs a rotation/reflection, so we just get a rotated unit circle. Second, $\Sigma$ scales along the coordinate axes, thus the circle gets deformed into an ellipse. And last, $U$ rotates the ellipse again.
In conclusion: the diagonal elements of $\Sigma$ are the lengths of the semi-axes, and the column vectors of $U$ give you the directions of the semi-axes. (If you are unfamiliar with computing the SVD, have a look at the link hereabove.)
A: Here's another way to approach this problem.
You can transform the given curve as follows
$ f(t) = \cos t \ u_1 + \sin t \ u_2 = \cos t' \ v_1 + \sin t' \ v_2 $
where $t' = t + h$ and $ v_1 \perp v_2 $.  To find $h$, plug in $ t = t' - h$ into $f(t)$.
$\begin{equation} \begin{split}
f(t) &= f(t' - h) = \cos(t' - h) \ u_1 + \ sin(t' - h) \ u_2 \\
& =\cos(t') ( \cos(h) \ u_1 - \sin(h) \ u_2 ) + \sin(t') ( \sin(h) \ u_1 + \cos (h) \ u_2 ) \\
& = \cos(t') v_1 + \sin(t') v_2 \end{split}\end{equation} $
Since we want $v_1 \perp v_2 $ then
$ ( \cos(h) \ u_1 - \sin(h) \ u_2 ) \cdot ( \sin(h) \ u_1 + \cos (h) \ 
 u_2 ) = 0 $
And this equation simplifies to
$ (u_1 \cdot \ u_2 ) ( \cos^2(h) - \sin^2(h) ) + \sin(h) \cos(h) (u_1 \cdot u_1 -  u_2 \cdot u_2) = 0$
which reduces to
$ (u_1 \cdot \ u_2 ) ( \cos(2h) ) + \dfrac{1}{2} \sin(2h) (u_1 \cdot u_1 -  u_2 \cdot u_2) = 0 $
Therefore,
$ \tan(2h) = \dfrac{ 2 u_1 \cdot u_2 }{ u_1 \cdot u_2 - u_2 \cdot u_2 } $
Now that we have $h$, vectors $v_1 $ and $v_2$ are fully specified.
Therefore, our problem now is to prove that
$ f(t') = \cos t' \ v_1 + \sin t' \ v_2 $
is an ellipse where it is known that $ v_1 $ and $v_2 $ are perpendicular to each other.  For that, define two perpendicular axes $X'$ and $Y'$ where the $X'$ axis is along $v_1$ and the $Y'$ axis is along $v_2$, i.e. the unit vector along the $X'$ axis is
$ \hat{X'} = \dfrac{ v_1 }{\| v_1 \| } $
and similarly the unit vector along the $Y'$ axis is
$ \hat{Y'} = \dfrac{v_2 }{ \| v_2 \| } $
Note that the axes $X'$ and $Y'$ are related to the standard $x$ and $y$ axes by a rotation (and a reflection if $u_2$ is clockwise from $u_1$).
Thus the vector $f$ is expressible as follows
$ f = ( \cos t' \| v_1 \| ) \hat{X'} + ( \sin t' \| v_2 \| ) \hat{Y'} $
Define the coordinates of $f$ with respect to the axis $X'$ and $Y'$ as $x'$ and $y'$ as follows
$ x' = \cos t' \| v_1 \| , \ y' = \sin t' \| v_2 \| $
Hence,
$ f = x' \hat{X'} + y' \hat{Y'} $
Now, it follows that $\cos t' = \dfrac{x'}{\| v_1 \|} , \sin t' = \dfrac{y'}{\|v_2 \|} $, and since $\cos^2 t' + \sin^2 t' = 1$ then
$ \dfrac{x'^2}{\| v_1 \|^2} + \dfrac{ y'^2 }{\| v_2 \|^2 } = 1 $
which is an equation of an ellipse.
A: The solution is already given nicely by other math lovers. I will try to do in my style. I changed OP's letters in my solution: $(a_1,a_2)=(a,b)$ and $(b_1,b_2)=(c,d)$.
Lets find the linear transformation $B: \Bbb{R}^2\rightarrow\Bbb{R}^2$ which sends $u_1=(a,b)$ to $e_1=(1,0)$ and $u_2=(c,d)$ to $e_2=(0,1)$. Sucha a map exists because, $u_1$ and $u_2$ are linearly independent. How can we find that map without solving a system of linear equations? By thinking the converse. The linear map which sends $e_1=(1,0)$ to $u_1=(a,b)$ and $e_2=(0,1)$ to $u_2=(c,d)$ is
$A(x,y)=(ax+cy,bx+dy)$. So,
$$B(x,y)=A^{-1}(x,y)=(\frac{dx-cy}{\Delta},\frac{-bx+ay}{\Delta})$$
where $\Delta=ad-bc\neq 0$.
After this transformation the new coordinates are $x'=\frac{dx-cy}{\Delta}$ and $y'=\frac{-bx+ay}{\Delta}$ and furthermore, we have $(x',y')=\cos te_1+\sin te_2=(\cos t,\sin t)$, that is, $(x')^2+(y')^2=1$. This equation, interms of old coordinates is
$$(b^2+d^2)x^2-2(ab+cd)xy+(a^2+c^2)y^2=(ad-bc)^2.$$
The discriminant of this conic is $\delta=(b^2+d^2)(a^2+c^2)-(ab+cd)^2=\Delta^2>0$, so it is an ellipse.
