TS EAMCET 04 May 2019 Shift 1 Solved Paper

x2+4xy+ky2−4x−10y+3=0
Then
|

1	2	−2
2	k	−5
−2	−5	3

|=0
(3k−25)−2(6−10)−2(−10+2k)=0
3k−25+8+20−4k=0
k=3
So, the equations of two lines are,
x+y−3=0‌ and ‌x+3y−1=0
The intersection points of the line are (4,-1) and distance from
‌ origin is ‌d=√17
The product of distance from origin to line is,
p‌‌=|‌

3

√2

×||‌

1

√10

|
‌‌=‌

3

2√5

This implies,
d2−20p2=17−9
=8