If the combined equation of the lines joining the origin to the points of intersection of the curve x2+y2−2x−4y+2=0 and the line x+y−2=0 is (l1x+m1y)(l2x+m2y)=0, then l1+l2+m1+m2=
Put x=2−y in the given conic ‌⇒(2−y)2+y2−2(2−y)−4y+2=0 ‌⇒y2−4y+4+y2−4+2y−4y+2=0 ‌⇒2y2−6y+2=0⇒y2−3y+1 ‌∴y=‌
3±√5
2
‌⇒y1=‌
3+√5
2
,y2=‌
3−√5
2
Thus, x1=2−(‌
3+√5
2
)=‌
1−√5
2
and x2=2−(‌
3−√5
2
)=‌
1+√5
2
then, p1≡(‌
1−√5
2
,‌
3+√5
2
) and p2≡(‌
1+√5
2
,‌
3−√5
2
) The lines OP1 and OP2 will have equation of the form ‌y=mx ‌⇒‌‌lx+my=0 ‌∵‌‌m1=‌
y1
x1
=‌
3+√5
1−√5
‌=‌
3+√5
1−√5
×‌
1+√5
1+√5
=−2−√5 ‌‌ and ‌m2=‌
y2
x2
=‌
3−√5
1+√5
‌=‌
3−√5
1+√5
×‌
1−√5
1−√5
=−2+√5 ∴ Combined equation of lines are (m1x−y1)(m2x−y)=0 will form ⇒(l1x+m1y)(l2x+m2y)=0 Lets expands and compare ‌m1m2x2−(m1+m2)xy+y2=l1l2x2 ‌+(l1m2+l2m1)xy+m1m2y2 ‌∴l1l2=m1m2,l1m2+l2m1=−(m1+m2), ‌m1m2=1 ‌m1=−2−√5,m2=−2+√5 ‌‌ and ‌m1+m2=−4 ‌∵‌‌l1l2=m1m2=1 ‌⇒‌‌l1=1∕l2 ‌∴‌‌l1m2+‌
m1
l1
=4 ‌⇒‌‌(√5−2l12+(−√5−2)=4l1. ‌⇒‌‌l1=1‌ and ‌l2=1 Hence, l1+l2+m1+m2=1+1−4=−2