#A(-a,0)# and #B(a,0)# are fixed points. C is a point which divides internally AB in a constant ratio #tan(alpha)#. If AC and BC subtend equal angles at P, prove that the equation of the locus of P is #x^2 + y^2 + 2ax sec(2alpha) + a^2=0#?

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Answer 1 · 2017-11-01T12:03:09

Let the coordinates of the movinng point #P # be #(h,k)#

Given that the coordinates of #A->(-a.0) and "of " B->(a,0)#

So the variable length of #AP=sqrt((h+a)^2+k^2)#

And the variable length of #BP=sqrt((h-a)^2+k^2)#

By the given condition AC and BC subtend equal angles at P. So PC must always be bisector of #/_APB#. This gives the relation

#(AP)/(BP)=(AC)/(BC)=tanalpha(given)#

So we have

#(sqrt((h+a)^2+k^2))/(sqrt((h-a)^2+k^2))=tanalpha#

#=>(sqrt((h-a)^2+k^2))/(sqrt((h+a)^2+k^2))=1/tanalpha#

#=>((h-a)^2+k^2)/((h+a)^2+k^2)=1/tan^2alpha#

By dividendo and componendo we get

#=>((h-a)^2-(h+a)^2+k^2-k^2)/((h-a)^2+(h+a)^2+2k^2)=(1-tan^2alpha)/(1+tan^2alpha)#

#=>(-4ha)/(2(h^2+a^2+k^2))=cos2alpha=1/(sec2alpha)#

#=>h^2+k^2+a^2=-2ahsec2alpha#

#=>h^2+k^2+2ahsec2alpha+a^2=0#

Inserting #h=x and k=y# we get the equation of the locus of point P as follows

#color(red)(=>x^2+y^2+2axsec2alpha+a^2=0)#