Identify the type of conic 4(x-2y+1)^2+9(2x+y+2)^2=254(x−2y+1)2+9(2x+y+2)2=25 ?

I reduced the equation to :
40x^2+30xy+9y^2+60x+36y+15=040x2+30xy+9y2+60x+36y+15=0
a=40, b=9, h=18a=40,b=9,h=18
How to do the next steps?

2 Answers
Shwetank Mauria
May 22, 2017

It is an ellipse.

Explanation:

Let the equation be of the type Ax^2+Bxy+Cy^2+Dx+Ey+F=0Ax2+Bxy+Cy2+Dx+Ey+F=0

then if

B^2-4AC=0B2−4AC=0 and A=0A=0 or C=0C=0, it is a parabola

B^2-4AC<0B2−4AC<0 and A=CA=C, it is a circle

B^2-4AC<0B2−4AC<0 and A!=CA≠C, it is an ellipse

B^2-4AC>0B2−4AC>0, it is a hyperbola

Now 4(x-2y+1)^2+9(2x+y+2)^2=254(x−2y+1)2+9(2x+y+2)2=25 can be written as

4(x^2+4y^2+1-4xy+2x-4y)+9(4x^2+y^2+4+4xy+4y+8x)=254(x2+4y2+1−4xy+2x−4y)+9(4x2+y2+4+4xy+4y+8x)=25

or 40x^2+20xy+25y^2+80x+20y+15=040x2+20xy+25y2+80x+20y+15=0

and we have A=40A=40, B=20B=20 and C=25C=25

and hence B^2-4AC=400-4000=-3600B2−4AC=400−4000=−3600

as B^2-4AC < 0B2−4AC<0 and A!=CA≠C, it is an ellipse.

graph{4(x-2y+1)^2+9(2x+y+2)^2=25 [-3.219, 1.78, -1.23, 1.27]}

Note if equation is 40x^2+30xy+9y^2+60x+36y+15=0#

we have B^2-4AC=900-1440=-540 and as A!=C,

this is also an ellipse but different one.

graph{40x^2+30xy+9y^2+60x+36y+15=0 [-9.64, 10.36, -6.78, 3.22]}

Cesareo R.
May 22, 2017

See below.

Explanation:

Making

{(u = x-2y+1),(v=2x+y+2):}

which is equivalent to a coordinate's change so we have in the new coordinates

4u^2+9v^2=25 or

(u/(5/2))^2+(v/(5/3))^2=1 which is the equation of an ellipse.

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