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

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5 months ago

```							 the calculations are insanely lengthy in this ques. but they can be simplified using parametric forms.let the line thru P beL: 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 Cso (w+2)^2= (mw+3)^2 + 2(w+2)(mw+3) rearrange to getw^2(m^2+2m – 1)+2w(5m+1)+17= 0this 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)= 17now 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)= ± 17or [17/(m^2+2m – 1)]*(1+m^2)= ± 17so (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 :))
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5 months ago
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