Conic sections

**pbeasley** · March 3rd 2009, 09:11 AM

5x2 + 6x - 4y = x2 - y2 -2x

**Soroban** · March 3rd 2009, 09:58 AM

Hello, pbeasley!

No instructions? . . . I'll have to guess . . .

$5x^2 + 6x - 4y \:= \:x^2 - y^2 -2x$

We have: . $4x^2 + 8x + y^2 - 4y \:=\:0 \quad\Rightarrow\quad 4(x^2 + 2x \qquad) + (y^2 - 4y \qquad) \:=\:0<br />$

Complete the square: . $4(x^2 + 2x {\color{blue}\:+\: 1}) + (y^2 - 4y {\color{red}\:+\: 4}) \:=\:0 {\color{blue}\:+\:4} {\color{red}\:+\: 4}$

And we have: . $4(x+1)^2 + (y-2)^2 \:=\:8 \quad\Rightarrow\quad \frac{(x+2)^2}{2} + \frac{(y-2)^2}{8} \:=\:1$

We have an ellipse.

. . Center: (-1,2)
. . Major axis (vertical): $a = 2\sqrt{2}$
. . Minor axis (horizontal): $b = \sqrt{2}$
. . Focal distance: $c = \sqrt{6}$
. . Eccentricity: $e = \tfrac{\sqrt{3}}{2}$

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