A line passing through the point P(2, 3) meets the lines represented by x^2-2xy -y^2=0 at the point A and B such that PA.PB = 17, the equation of the line is 1) x = 2 2) x=5 3)3x-2y=0 4) y=3

							
 
the calculations are insanely lengthy in this ques. but they can be simplified using parametric forms.
let the line thru P be
L: y – 3= m(x – 2)
or (y – 3)/m = x – 2= w (say)
so, any pt on L will be of the form (w+2, mw+3)
now, to find intersection of L and pair of st lines C: x^2-2xy -y^2=0, (w+2, mw+3) would also lie on C
so (w+2)^2= (mw+3)^2 + 2(w+2)(mw+3) 
rearrange to get
w^2(m^2+2m – 1)+2w(5m+1)+17= 0
this will obviously have max 2 real roots of w, call them w1 and w2.
these pts of intersection would hence be A(w1+2, mw1+3) and B(w2+2, mw2+3)
now PA= sqrt[(w1+2 – 2)^2 + (mw1+3 – 3)^2] and PB= sqrt[(w2+2 – 2)^2 + (mw2+3 – 3)^2] 
PA= |w1|sqrt(1+m^2) and PB= |w2|sqrt(1+m^2) 
so PA.PB= |w1*w2|(1+m^2)= 17
now from w^2(m^2+2m – 1)+2w(5m+1)+17= 0 or w1w2= 17/(m^2+2m – 1)
hence  |w1*w2|(1+m^2)= 17 implies  w1*w2(1+m^2)= ± 17
or [17/(m^2+2m – 1)]*(1+m^2)= ± 17
so (1+m^2)= ± (m^2+2m – 1)
(1+m^2)= + (m^2+2m – 1) or (1+m^2)= – (m^2+2m – 1)
now comes a VERY IMPORTANT concept: whenever a quad eqn reduces to a linear eqn (through cancellations), one of the roots has got to be infinity (the proof i cannot cover here).
so, roots of (1+m^2)= + (m^2+2m – 1) are m= 1, infinity.
the roots of (1+m^2)= – (m^2+2m – 1) are m= 0, – 1.
so, the 4 lines are of the form y – 3= m(x – 2) corresponding to these 4 values of m.
lines are y= 3 (m=0)
x= 2 (m= infinity)
y= x+1 (m=1)
y+x= 5 (m= – 1)
all these lines are correct (none of them is rejected). Hence this forms the complete set of solns.
in options only (1) and (4) are correct.
KINDLY APPROVE :))