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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) y = 3 3)3x-2y=0 4) x = 5

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) y = 3
3)3x-2y=0
4) x = 5

Question Image
Grade:12th pass

3 Answers

Saurabh Koranglekar
askIITians Faculty 10341 Points
one year ago
Dear student

Let the equation of required line be y = mx+c passing through (2,3)

The pair of lines can be separated as

y = 0.4142 x let this have A ( x1,y1)

y = -2.4142 x let this have B (x2,y2)


Where A,B lie on the line y = mx +c

Now PA*PB = 17

Using the equations m and c can be determined

And thus the answer

Please tell if the doubt persists

Regards


Aditya Gupta
2075 Points
one year ago
 
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 (2) are correct.
KINDLY APPROVE :))
Aditya Gupta
2075 Points
one year ago
NOTE that in your pic only option 2 is marked correct.
but IT IS WRONG.
both options 1 and 2 will be CORRECT.
I have checked it on desmos graphing calc too.

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