I am trying to put the equations into standard form of a conic section...

But cannot find its factors properly i have those three equations left

Show all steps please...

Thanks

1. 3x^2-y^2+9x+8y-145/4=0

2. 4x^2+9y^2+24x+18y+12=0

3. y^2+10y=2x-1

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- Nov 12th 2007, 02:30 PMxterminal01[SOLVED] Equation into standard form of a conic
I am trying to put the equations into standard form of a conic section...

But cannot find its factors properly i have those three equations left

Show all steps please...

Thanks

1. 3x^2-y^2+9x+8y-145/4=0

2. 4x^2+9y^2+24x+18y+12=0

3. y^2+10y=2x-1

- Nov 12th 2007, 07:31 PMtopsquark
$\displaystyle 3x^2 - y^2 + 9x + 8y -\frac{145}{4} = 0$

Let's get rid of that nasty fraction:

$\displaystyle 12x^2 - 4y^2 + 36x + 32y - 145 = 0$

Complete the square for x and y:

$\displaystyle (12x^2 + 36x) + (-4y^2 + 32y) = 145$

$\displaystyle 12(x^2 + 3x) - 4(y^2 - 8y) = 145$

$\displaystyle 12 \left ( x^2 + 3x + \frac{9}{4} \right ) - 4 (y^2 - 8y + 16) = 145 + 12 \cdot \frac{9}{4} - 4 \cdot 16$

$\displaystyle 12 \left ( x + \frac{3}{2} \right ) ^2 - 4(y - 4)^2 = 108$

$\displaystyle \frac{ \left ( x + \frac{3}{2} \right ) ^2}{\frac{108}{12}} - \frac{(y - 4)^2}{\frac{108}{4}} = 1$

$\displaystyle \frac{ \left ( x + \frac{3}{2} \right ) ^2}{3^2} - \frac{(y - 4)^2}{(\sqrt{27})^2} = 1$

which is the form for a hyperbola.

The rest are done in a similar fashion.

-Dan - Nov 12th 2007, 10:25 PMxterminal01
Thank you I appreciate it,

I have tried to do the # 2 and 3 myself and this is what i came up with

2. 4x^2+9y^2+24x+18y+12=0

4x^2+9y^2+24x+18y=-12

4(x^2+6x+9)+9(y^2+2y+1)=-12+36+9

4(x+3)^2 + 9(y+1)^2 = 33

(x+3)^2 + (y+1)^2

33 33 = 1

4 9

and # 3 i guess is a parabola and this is what i got

y^2+10y=2x-1

(y^2+10y+25) = 2x-1+25

(y+5)^2 = 2x + 24

(y+5)^2 = 2(x+12)

Please let me know if i did the rest correct :) - Nov 12th 2007, 11:23 PMearboth