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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.

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  • $\begingroup$ May be projective geometry has an elegant solution among parallel sections after locating the original cone apex projection. $\endgroup$ – Narasimham Feb 5 '18 at 18:27
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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,

$$\color{blue}{F'}= \frac{1} {\det \begin{pmatrix} A & \frac{B}{2} \\ \frac{B}{2} & C \\ \end{pmatrix}} \left[ \frac{AE^2+CD^2-BDE}{4}+p^2 \det \begin{pmatrix} A & \frac{B}{2} & \frac{D}{2} \\ \frac{B}{2} & C & \frac{E}{2} \\ \frac{D}{2} & \frac{E}{2} & F \end{pmatrix} \right] $$

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  • $\begingroup$ What are $a,b,c,d,e$? And I think you mean $p^2$ not $\sqrt{p}$. $\endgroup$ – Jan-Magnus Økland Feb 5 '18 at 18:52
  • $\begingroup$ @Jan-MagnusØkland Answer revised. $\endgroup$ – Ng Chung Tak Feb 5 '18 at 19:03
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You obtain this effect by rescaling the coordinate axis by the factor $p$, and the equation becomes

$$ A\frac{x^2}{p^2} + B\frac{xy}{p^2} + C\frac{y^2}{p^2} + D\frac{x}{p} + E\frac{y}{p} + F =0. $$

If the center must remain unchanged, translate the center to the origin (the center is found by solving $2Ax+By+F=0,Cx+2Dy+E=0$), dilate and translate back.

The combined transform is

$$x\to\frac{x-x_c}p+x_c,\\y\to\frac{y-y_c}p+y_c.$$

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