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