Stretching an ellipse along major or minor axis Consider the ellipse given by:
$$
Ax^2 + Bxy + Cy^2 + Dx + Ey + F =0.
$$
What is the equation of an ellipse which has major and minor axis equal to $p$ times the major and minor axis length of the above ellipse.
My attempt is as follows:
We can remove rotation, increase axis length and then rotate back. An example of rotation is given below:
Rotating a conic section to eliminate the $xy$ term.
I am wondering if there is less complicated intuition into this problem or less complicated way.

Referring to the standard results here, the centre is given by 
$$(h,k)=
\left(
  \frac{2CD-BE}{B^2-4AC}, \frac{2AE-BD}{B^2-4AC}
\right)$$
and the transformed conics is
$$\frac{A+C \color{red}{\pm} \sqrt{(A-C)^{2}+B^{2}}}{2} X^2+
\frac{A+C \color{red}{\mp} \sqrt{(A-C)^{2}+B^{2}}}{2} Y^2+
\frac
{\det
  \begin{pmatrix}
    A & \frac{B}{2} & \frac{D}{2} \\
    \frac{B}{2} & C & \frac{E}{2} \\
    \frac{D}{2} & \frac{E}{2} & F
  \end{pmatrix}}
{\det
  \begin{pmatrix}
    A & \frac{B}{2} \\
    \frac{B}{2} & C \\
  \end{pmatrix}}=0$$

It's just simply re-scaling the constant term $F$, that is 
$$Ax^2+Bxy+Cy^2+Dx+Ey+\color{blue}{F'}=0$$ where
$$p^2
\det
\begin{pmatrix}
  A & \frac{B}{2} & \frac{D}{2} \\
  \frac{B}{2} & C & \frac{E}{2} \\
  \frac{D}{2} & \frac{E}{2} & F
\end{pmatrix}
=
\det
\begin{pmatrix}
  A & \frac{B}{2} & \frac{D}{2} \\
  \frac{B}{2} & C & \frac{E}{2} \\
  \frac{D}{2} & \frac{E}{2} & \color{blue}{F'}
\end{pmatrix}$$
On solving,
$$\color{blue}{F'}=
\frac{1}
     {\det
      \begin{pmatrix}
        A & \frac{B}{2} \\
        \frac{B}{2} & C \\
      \end{pmatrix}}
\left[
  \frac{AE^2+CD^2-BDE}{4}+p^2
      \det
      \begin{pmatrix}
        A & \frac{B}{2} & \frac{D}{2} \\
        \frac{B}{2} & C & \frac{E}{2} \\
        \frac{D}{2} & \frac{E}{2} & F
      \end{pmatrix}
\right]
$$

