# [SOLVED] Equation into standard form of a conic

• Nov 12th 2007, 02:30 PM
xterminal01
[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
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 PM
topsquark
Quote:

Originally Posted by xterminal01
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
Thanks

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

$3x^2 - y^2 + 9x + 8y -\frac{145}{4} = 0$

Let's get rid of that nasty fraction:
$12x^2 - 4y^2 + 36x + 32y - 145 = 0$

Complete the square for x and y:
$(12x^2 + 36x) + (-4y^2 + 32y) = 145$

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

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

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

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

$\frac{ \left ( x + \frac{3}{2} \right ) ^2}{3^2} - \frac{(y - 4)^2}{(\sqrt{27})^2} = 1$
which is the form for a hyperbola.

The rest are done in a similar fashion.

-Dan
• Nov 12th 2007, 10:25 PM
xterminal01
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 PM
earboth
Quote:

Originally Posted by xterminal01
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
(how about some squares as denominator?)

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 :)

Hi,

(And now it's time to learn a little bit $L_A T^E \chi$)