×

Hello! Socratic's Terms of Service and Privacy Policy have been updated, which will be automatically effective on October 6, 2018. Please contact hello@socratic.com with any questions.

How do you use rotation of axes to identify and sketch the curve of #97x^2+192xy+153y^2=225#?

1 Answer
Feb 15, 2018

Answer:

The curve is an ellipse and after rotation it is #9x^2+y^2=9#

Explanation:

A conic equation of the type of #Ax^2+Bxy+Cy^2+Dx+Ey+F=0# is rotated by an angle #theta#, to form a new Cartesian plane with coordinates #(x',y')#, if #theta# is appropriately chosen, we can have a new equation without term #xy# i.e. of standard form.
http://philschatz.com/precalculus-book/contents/m49441.html
The relation between coordinates #(x,y)# and #(x'.y')# can be expressed as
#x=x'costheta-y'sintheta# and #y=x'sintheta+y'costheta#

or #x'=xcostheta+ysintheta# and #y=-xsintheta+ycostheta#

for this we need to have #theta# given by #cot2theta=(A-C)/B#

In the given case as equation is #97x^2+192xy+153y^2-225=0#, we have #A-C=-56# and #B=192# and hence #cot2theta=-7/24# i.e. #(cot^2theta-1)/(2cottheta)=-7/24# or #12cot^2theta+7cottheta-12=0# or #(4cottheta-3)(3cottheta+4)=0# i.e. #cottheta=3/4# or #cottheta=-4/3#.

We consider only acute angle i.e. #cottheta=3/4#, which leads to #costheta=3/5# and #sintheta=4/5#

Hence relation is give by #x=3/5x'-4/5y'# and #y=4/5x'+3/5y'#

Hence, we get #97((3x')/5-(4y')/5)^2+192((3x')/5-(4y')/5)((4x')/5+(3y')/5)+153((4x')/5+(3y')/5)^2-225=0#

or #97((9x'^2)/25+(16y'^2)/25-(24x'y')/25)+192((12x'^2)/25-(12y'^2)/25-(7x'y')/25)+153((16x'^2)/25+(9y'^2)/25+(24x'y')/25)-225=0#

or #225x'^2+25y'^2-225=0# or #9x'^2+1y'^2=9#

The two graphs are as follows:
graph{97x^2+192xy+153y^2-225=0 [-10, 10, -5, 5]}
and
graph{9x^2+y^2=9 [-10, 10, -5, 5]}