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Trace the conic? 8x^2+4xy+5y^2=24(x+y)


Sneh Sajwan 29th Nov, 2019
Answer (1)
Avishek Dutta 29th Nov, 2019
ax^2+2hxy+by^2+2gx+2fy+c=0.
8x^2+4xy+5y^2-24x-24y=0
a=8,b=5,c=0,h=2,f=g=-12
D=abc+2fgh-af^2-bg^2-ch^2
=8*5*0+2*(-12)(-12)*2-8*(-12)^2-5*(-12)^2-0*2^2
=-1296
Now
If D=0

If h^2-ab > 0,the equation represents two distinct real lines.
If h^2-ab = 0,the equation represents parallel lines.
If h^2-ab < 0,the equation represents non-real lines.

If D is non zero

If h^2-ab>0,it represents a hyperbola and a rectangular hyperbola(a+b=0).
If h^2-ab=0,the equation represents a parabola.
If h^2-ab<0,the equation represents a circle(a=b, h= 0) or an ellipse(a not equal to b).(For a real ellipse,
D/(a+b)<0).
Here D is non zero and h^2-ab=2-58=-36.
And also a not equal to b.Hence it is an ellipse.
Also D/(a+b)<0.
It is a real ellipse.
1 Comment
Comments (1)
29th Nov, 2019
thank u so much. Can you find the centre and length of axes
Reply

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