09-09-2015, 04:11 PM
I would like to know if there is a better way to do this on the HP prime. Is there a built in function or something I can use to simplify this process, besides writing a custom program? It's so tedious.
Transform the equation in x and y into an equation in X and Y (without an XY-term) by rotating the x- and y-axes by the indicated angle. I am using the Prime to verify my answers.
Problem: y^2 -sqrt(3)*x*y + 3 = 0 where theta is 30 degrees.
For reference: x=x, y=y, a=X', b=Y' and c=theta
[CAS]
[set degree mode]
x:=a*cos(c)-b*sin(c)
y:=a*sin(c)+b*cos(c)
c:=30
then use the stored variables to evaluate the equation that I'm working with.
y^2-sqrt(3)*x*y+3=0
That input simplifies to
[(-a)^2+3b^2+6]/2=0
Takes a little more manual transformation to get it into the standard form of a hyperbola, and I arrive then at the correct answer:
[(a^2)/6] -[(b^2)/2] = 1 where a = x and b=y.
Transform the equation in x and y into an equation in X and Y (without an XY-term) by rotating the x- and y-axes by the indicated angle. I am using the Prime to verify my answers.
Problem: y^2 -sqrt(3)*x*y + 3 = 0 where theta is 30 degrees.
For reference: x=x, y=y, a=X', b=Y' and c=theta
[CAS]
[set degree mode]
x:=a*cos(c)-b*sin(c)
y:=a*sin(c)+b*cos(c)
c:=30
then use the stored variables to evaluate the equation that I'm working with.
y^2-sqrt(3)*x*y+3=0
That input simplifies to
[(-a)^2+3b^2+6]/2=0
Takes a little more manual transformation to get it into the standard form of a hyperbola, and I arrive then at the correct answer:
[(a^2)/6] -[(b^2)/2] = 1 where a = x and b=y.