Answer to Question #102642 in Linear Algebra for BIVEK SAH

Question

Identify the conic section whose equation is

5x^2-4xy+5y^2=4

Answer 1

Answer: Ellipse

$5x^2 - 4xy + 5y^2 = 4$

If we draw a graph of this function we'll get:

Proof:

$\begin{Bmatrix} x=x_1cos(\phi)-y_1sin(\phi) \\ y=x_1sin(\phi)+y_1cos(\phi) \end{Bmatrix}$

$5(x_1cos(\phi)-y_1sin(\phi))^2-4(x_1cos(\phi)-y_1sin(\phi))(x_1sin(\phi)+y_1cos(\phi))+5(x_1sin(\phi)+y_1cos(\phi))^2=4$

After simplifying we will write only expression with $x_1y_1$ and this expression equal to zero:

$4(x_1y_1cos^2(\phi) - x_1y_1sin^2(\phi))=0$

$\cos^2(\phi)-\sin^2(\phi)=0;\\\phi=\pi/4;$

$\begin{Bmatrix} x=x_1cos(\pi/4)-y_1sin(\pi/4) =\frac{x_1}{\sqrt{2}} - \frac{y_1}{\sqrt{2}} \\ y=x_1sin(\pi/4)+y_1cos(\pi/4)=\frac{x_1}{\sqrt{2}} + \frac{y_1}{\sqrt{2}} \end{Bmatrix}$

and we will get:

$\frac{3x_1^2}{4} + \frac{7y_1^2}{4} =1$

This is an equation of the ellipse.