which conic is represented by 11x²-20xy+10y²+4x+8y+3=0
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Given conic S:2x2−72xy+23y2−4x−28y−48=0
The general equation of a conic section is a second-degree can be written as
f(x,y)=ax2+2hxy+by2+2gx+2fy+c
Here, h=−36,a=2,b=23,g=−2,f=−14,c=−48
△=abc+2fgh−af2−bg2−ch2
=(2)(23)(−48)+2(−14)(−2)(−36)−2(−14)2−23(−2)2−(−48)(−36)2
=−2208−2016−392−92+62208
=59300=0
h2−ab=(−36)2−2×23=1250
i.e. h2−ab>0
So, it represents a hyperbola.
Now, ∂x∂S=4x−72y−4
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