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Analytical Geometry

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25 Oct 2007 14:02:47 IST

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1)A and B are two fixed pts on the x-axis where OA=a,OB=2a.P is any pt in the x-y plane . AP andBP meet the y-axis at C and D resp. ADcuts OP at Q prive that CQ passes through a fixed for all position of P

2)A variable line through the pt P(-1,2) cuts the axis at A&B resp.Q is a pt on AB such that PQ is the HM(harmonic mean) of PA & PB Show that the locus ofQ is the line y=2x

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Bipin Dubey

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Joined: 23 Jan 2007

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25 Oct 2007 14:34:31 IST

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Let the variable point P be (h,k). It is known A = (a,0) and B = (2a,0)

Eqn of AP : y = (k/(h-a)) .(x-a)
It intersects y-axis at C where x=0, C = {0, -ak(h-a)}

Eqn. of BP : y = (k/(h-2a)) .(x-2a)
It intersects y-axis at D where x=0, D = {0, -2ak(h-2a)}

Eqn. of line through A(a,0) and D{0, -2ak(h-2a)}
AD : y = (2k/(h-2a)).(x-a)

Eqn. of OP : y = (k/h)x

Solve AD and OP to find their intersection i.e. Q.
You'll get Q = { 2ah/(h+2a) , 2ak/(h+2a) }

Also we know C = {0, -ak(h-a)}

Slope of CQ = {2ak/(h+2a) + ak(h-a)} / {2ah/(h+2a)} = 3k/2(h-a)

So equation of CQ : y + ak(h-a) = {3k/2(h-a)}.x

CQ : y - {3k/2(h-a)}.(x - 2a/3) = 0

which clearly passes through (2a,3 , 0).

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