Can someone please help me with this,
I am suppose to state which type of conic section is represented by each equation step by step if its possible
1. x^2-6x+y=8
2. 3x^2+5y^2+6x-10y=16
3. 2x^2+8x=2y^2-y+10
4. 3x^2+x-y^2+y=12
5. x^2+4y^2=8
Can someone please help me with this,
I am suppose to state which type of conic section is represented by each equation step by step if its possible
1. x^2-6x+y=8
2. 3x^2+5y^2+6x-10y=16
3. 2x^2+8x=2y^2-y+10
4. 3x^2+x-y^2+y=12
5. x^2+4y^2=8
Hello,
rearrange your equations until you can determine the type of conic. All of them have their axes parallel to the coordinate axes so completing the square will do:
to #1:
$\displaystyle x^2-6x+y = 8~\iff~(x^2-6x+9)+y=8+9~\iff~ y = -(x-3)^2+17$
That's a parabola opening down.
to #2:
$\displaystyle
3x^2+5y^2+6x-10y=16~\iff~3(x^2+2x+1) + 5(y^2-2y+1)=16+3+5~\iff~$ $\displaystyle 3(x+1)^2+5(y-1)^2=24$ . Now divide by 24 and you'll get:
$\displaystyle \frac{(x+1)^2}{8}+\frac{(y-1)^2}{\frac{24}{5}}=1$ That's the equation of an ellipse.
The ##3 to 5 should be done similary.(H, H, E)
Hi,
I was quite sure that you could do the last problems using the way I demonstrated to you, but .... I'll start the problems and I'll leave the final brush up to you:
to #3:
$\displaystyle 2x^2+8x=2y^2-y+10~\iff~2x^2+8x-2y^2+y=10~\iff~$ $\displaystyle 2(x^2+4x+4)-2(y^2+\frac12 y+\frac{1}{16})= 10+8-\frac{2}{16}$ . Simplify and you should come out with a hyperbola
to #4:
$\displaystyle 3x^2+x-y^2+y=12~\iff~3(x^2+\frac13 x+\frac{1}{36})-(y^2-y+\frac14)=12 + \frac{3}{36} - \frac14$ Simplify. It's a hyperbola too.
to #5:
$\displaystyle x^2+4y^2=8$ Divide by 8. Then you get the equation of an ellipse.
Hello, xterminal01!
If we are to identity the type of conic only , there is an "eyeball" method.State which type of conic section is represented by each equation.
$\displaystyle 1)\;\;x^2-6x+y\:=\:8$
$\displaystyle 2)\;\;3x^2+5y^2+6x-10y\:=\:16$
$\displaystyle 3)\;\;2x^2+8x\:=\:2y^2-y+10\quad\Rightarrow\quad 2x^2+8x-2y^2+y\:=\:10$
$\displaystyle 4)\;\;3x^2+x-y^2+y\:=\:12$
$\displaystyle 5)\;\;x^2+4y^2\:=\:8$
First, get all the variables on one side of the equation.
. . (As we did in #3.)
If it has either $\displaystyle x^2$ or $\displaystyle y^2$, but not both, it is a parabola. .(#1)
If $\displaystyle x^2$ and $\displaystyle y^2$ have the same coefficient, it is a circle. .(None listed)
If $\displaystyle x^2$ and $\displaystyle y^2$ have the same sign, it is an ellipse. .(#2 and 5)
If $\displaystyle x^2$ and $\displaystyle y^2$ have opposite signs, it is a hyperbola. .(#3 and 4)
Hello,
you've done #5 correctly. I would have written:
$\displaystyle \frac{x^2}{(2\sqrt{2})^2}+\frac{y^2}{(\sqrt{2})^2} =1$
#4: I'll continue:
$\displaystyle 3(x^2+\frac13 x+\frac{1}{36})-(y^2-y+\frac14)= \frac{71}{6}~\iff~ 3\left(x+\frac16\right)^2-\left(y-\frac12\right)^2=\frac{71}{6}$ and now divide by $\displaystyle \frac{71}{6}$. That's all.
#3: Divide by $\displaystyle \frac{143}{8}$ to get a 1 at the RHS.