TS EAMCET 19-July-2022 Shift 2 Solved Paper

Given, the equation of line 4x−6y−2=0 and the equation of curve

3x2−4xy+5y2−2x+y−6=0
The equation of line 4x−6y−2=0 can be written as 2x−3y=1
The equation of curve can be written as
(3x2−4xy+5y2)−(2x−y)(1)−6(1)2=0
Now, substitute 1=2x−3y in equation of curve
‌⇒3x2−4xy+5y2−4x2+8xy−3y2
‌‌−24x2−54y2+72xy=0
‌⇒‌‌−25x2+76xy−52y2=0
‌f(x,y)=25x2−76xy+52y2
‌f(1,−1)=25+76+52=153
‌f(−1,−1)=25−76+52=1
‌∴‌‌‌

f(1,−1)

f(−1,−1)

=153