AP EAMCET Engineering 2018 Apr 22 Shift 2 Paper

The equation,
3x2+7xy+2y2+2gx+2fy+2=0
Above equations shows the pair of intersecting lines, so1 (3)(2)2+7gf−3f2−2g2−49=0
24+114gf−6f2−4g2−49=0
4g2+6f2−14gf+25=0
To get point of interaction of pair of straight line the equation should be differentiate with respect to x and y .
So,
6x+7y+2g=0 …… (I)
7x+4y+2f=0 …… (II)
Now, point of interaction of line (I) and (II) is,
(

14f−8g

25

,

14g−12f

25

)
From question,
(

14f−8g

25

)2+(

14g−12f

25

)2=

2

5

[196f2+64g2−224fg+196g2+144f2 −336gf]=250
26g2+34f2−56gf=25
Now,
26g2+34f2−56gf=25
16g2+24f2−56gf=−100
−‌‌‌‌−‌‌‌‌‌‌+‌‌‌‌‌‌‌‌‌‌+
––––––––––––––––––––––––
10g2+10f2=125
f2+g2=

25

2