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If I have a conic in standard form (converted from 9x2 -4y2-36x-24y=0) of 9(x-2)2-4(y+3)2=0, how would I go about either solving or graphing this. If it were a plus sign between them I'd just give an answer of a point, but I have a strange feeling that there is more to the answer, a solution with an explanation would be greatly appreciated.
this is a degenerate conic ...
http://youtu.be/ADLovwJA0Is
Hello, mikewezyk!
Quote:
$\displaystyle \text{If I have a conic in standard form }\,9(x-2)^2 - 4(y+3)^2 \:=\:0$
. . $\displaystyle \text{ converted from }\,9x^2 - 4y^2 - 36x - 24y \:=\:0$
$\displaystyle \text{how would I go about either solving or graphing this?}$
If it were a plus sign between them, I'd just give an answer of a point, . Right!
. . but I have a strange feeling that there is more to the answer. . Right again!
A solution with an explanation would be greatly appreciated.
We have: .$\displaystyle 9(x-2)^2 - 4(y+3)^2 \:=\:0 \quad\Rightarrow\quad 4(y+3)^2 \:=\:9(x-2)^2$
. . . . . . . . $\displaystyle (y+3)^2 \:=\:\tfrac{9}{4}(x-2)^2 \quad\Rightarrow\quad y + 3 \:=\:\pm\tfrac{3}{2}(x-2) $
. . . . . . . . $\displaystyle y \;=\;-3 \pm\tfrac{3}{2}(x-2)$
These are the equations of the asymptotes of the hyperbola.
