Rotating axis

[kaebun]

identify the type of conic, solve for y, and graph the conic. Approximate the angle of rotation to eliminate the xy term

\(\displaystyle 4x^2\,-\, 6xy\,+\,2y^2\, -3x\,+\,10y\,-\,6=0\)

I was wondering if some one could tell me if the following is the right/ only way to do this problem, my teacher started it on the board and what she did make no sence to me and it doesn't even seem (at least to me) to be moving in the direction to answer the question.

\(\displaystyle cot2a\, =\, \frac{4\, -\, 2}{-6}\, =\,\frac{1}{-3}\)

\(\displaystyle tan2a \,=\, \frac{-3}{1}\)

\(\displaystyle sin2a\, =\,\frac{-3}{\sqrt{10}}\, \rightarrow\, \sqrt{\frac{1\,-\,\left(\frac{3}{\sqrt{10}}\right)}{2}}\, =\, \sin{(a)}\)

\(\displaystyle cos2a\,=\, \frac{1}{\sqrt{10}}\, \rightarrow\, \sqrt{\frac{1\,-\,\left(\frac{1}{\sqrt{10}}\right)}{2}} \,= \,\cos{(a)}\)

\(\displaystyle \sin{(a)}\, =\, \sqrt{0.348}\)

\(\displaystyle cos{(a)}\, =\, \sqrt{0.658}\)

if this is right could some one explain why.... thank you

 

[skeeter]

this page may help.

 

[kaebun]

thank you, i understand the concepts. Its just that the way i understand this problem is that i am supposed to name the conic, a ellipse i think, solve for y which could e done w/out rotating the axis and then fin the angle of rotation which would be \(\displaystyle cot(2a)= (4-2)/-6
cot 2a= -1/3
tan 2a= -3
2a= tan^-1 (-3)
a= -35.8\)
and this is not what she is doing at all and so i am confused :?
i don't think what she was doing answers the question

identify the type of conic, solve for y, and graph the conic. Approximate the angle of rotation to eliminate the xy term

but im not sure because im not sure what she is doing at all

 

[soroban]

Hello, kaebun!

Identify the type of conic, solve for y, and graph the conic.
Approximate the angle of rotation to eliminate the xy term

\(\displaystyle 4x^2\,-\, 6xy\,+\,2y^2\, -3x\,+\,10y\,-\,6\;=\;0\)

What you wrote is correct, but some of it is not necessary.

Given: \(\displaystyle \,Ax^2\,+\,Bxy\,+\,Cy^2\,+\,Dx\,+\,Ey\,+\,F\;=\;0\)
\(\displaystyle \;\;\)Angle of rotation, \(\displaystyle \theta:\;\;\tan2\theta\:=\:\frac{A\,-\,C}{B}\)

You have: \(\displaystyle \,\tan2\theta\:=\:\frac{-6}{4\,-\,2}\:=\:-3\)
Then: \(\displaystyle 2\theta\:=\:\tan^{-1}(-3)\:=\:71.565252559^o\;\;\Rightarrow\;\; \theta\:\approx\:-35.8^o\)

The discriminant is: \(\displaystyle \,D\:=\:B^2\,-\,4AC\)
\(\displaystyle \;\;\)You have: \(\displaystyle D\:=\-6)^2\,-\,4(4)(2)\:=\:+4\,\cdots\,\text{hyperbola}\)

Did it really say "solve for y" ?

 

[kaebun]

Did it really say "solve for y" ?

yep, you have to solve for to put it in you graphing calc and graph it.
To solve for y you do not have to rotate the axis, right?