Question

Rotation of a Hyperbola
Rotate the axes to eliminate the xy term and write the equation xy + 2 = 0 in standard form and sketch its graph.

Answer 1

let
X = xcos(A) + y Sin(A)
Y = ycos(A) - xSin(A)

find the value of A such that xy term vanishes.

Answer 2

what is A?

Answer 3

A is an angle of rotation.

Answer 4

put those values in x and y,
multiply them, can separate XY term, and equate the coefficients of xy ... and get the value of A

Answer 5

so what would i put into x and y?
and what does the X and Y mean are thet different from the lower case x and y?

Answer 6

yes big X and big Y

Answer 7

hold up what do i put into x and y?

Answer 8

(Xcos(A) + Y Sin(A))(Ycos(A) - XSin(A)) + 2 = 0

Answer 9

im so sorry for asking dumb questions but i still dont get what i put into the X and Y...

Answer 10

just multiply the above expression and simplify.

Answer 11

you have already put ... not you but me.
x*y+2 = 0
(Xcos(A) + Y Sin(A))*y + 2 = 0;
(Xcos(A) + Y Sin(A))(Ycos(A) - XSin(A)) + 2 = 0;

Answer 12

how do i solve for X, Y or A?

Answer 13

you don't need to solve your X,Y
Just solve for A, equating coefficinets of XY to zero.

Answer 14

so i simplified it and got XY(cos(A)) - X^2(cos(A))(sin(A)) + y^2(sin(A))(cos(A)) - XY(sin(A))^2= -2
what do i do now?

Answer 15

coeffiecitnt of XY must be zero ... from there get the value of A

Answer 16

-X^2(cos(A))(sin(A)) + y^2(sin(A))(cos(A))= -2
so would it look like this then?
if it does then i dont know where to go from here.....sorry for bothering you so much.. :)

Answer 17

no get A by equationg the coefficein of XY = zero

Answer 18

how would you do that?

Answer 19

where did you put your xy terms??

Answer 20

i thought they were zero so i took then off cuase that what i thought you ment...

Answer 21

okay so starting over did i do this equation right?
XY(cos(A)) - X^2(cos(A))(sin(A)) + y^2(sin(A))(cos(A)) - XY(sin(A))^2= -2

Answer 22

XY(cos(A)) - XY(sin(A))^2 <--- it's coefficient must be zero.

Answer 23

aight THANKS SO MUCH i think the standard form would be (y/4)-(x/4)=1and its tilted at a 45 degree angle....

Answer 24

well, that is a line, not a hyperbola ... and your original curve was rectangular huperbola.

Answer 25

anyway welcome.

Answer 26

no thts standard form of a hyperbola....its definity not a line