1) A line drawn from P(p,q) intersects the curve x^2+2hxy+by^2-1=0 at points Q and R(P not lying on th curve).If the product PQ . PR is always equal to 2,then:
a)p=1, q=-1
b)p^2 + q^2 = 2
c)p=0, q=2
d)p=-1, q=1
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UP 0 DOWN 0 0 2
2 Answers
62let the line be
(x-p) = m(y-q)
then put x= p+m(y-q) to get
\left( p+m(y-q)\right)^2+2h(p+m(y-q))y+by^2-1 \\=(p-mq)^2+m^2y^2-2(p-mq)my+2h(p-mq)y+2hmy^2+by^2-1 \\=y^2(m^2+2hm+b)+2y(hp-hmq-mp+m^2q)+(p-mq)^2-1 \\=y^2(m^2+2hm+b)+2y(h-m)(p-mq)+(p-mq)^2-1
we have to now find (y2-q).(y1-q) this will be in some way related to L2L1
the reason is that the slope is there.. we know the change in y coordinate..
hence the length will be (y2-q).(y1-q) (1+m)2/m2
Am too lazy to solve this... but I am sure you can pull this one off on a paper ...