How do use the method of translation of axes to sketch the curve #16x^2+9y^2-320x-108y+1780=0#?

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Answer 1 · 2017-12-10T08:38:48

Please refer to the Discussion in the Explanation.

The eqn. of the curve is

#S : 16x^2+9y^2-320x-108y+1780=0.#

#:. 16x^2-320x+9y^2-108y=-1780.#

#:. 16(x^2-20x)+9(y^2-12y)=-1780.#

Completing the squares, we have,

#16{(x^2-20x+100)-100}+9{(y^2-12y+36)-36}=-1780.#

#:. 16{(x-10)^2-100}+9{(y-6)^2-36}=-1780.#

#:. 16(x-10)^2-1600+9(y-6)^2-324=-1780.#

#:. 16(x-10)^2+9(y-6)^2=1600+324-1780=144.#

Dividing by #144,# we have,

# S : (x-10)^2/9+(y-6)^2/16=1.#

Now, shifting the Origin to the point #(10,6),# suppose that,

#(x,y)# becomes #(X,Y).#

#:. x=X+10, y=Y+6.#

#rArr(x-10)=X, and (y-6)=Y.#

#:. S : X^2/9+Y^2/16=1.#

This represents an ellipse with centre at #(0,0)# in #(X,Y)#

system and having lengths of major and minor axes #8 and 6,#

resp.