Question No 1.
Let the coordinate of the moving point P be #(h,k)#
The distance of P from #(-3,0)#
#d_1=sqrt((h+3)^2+(k-0)^2)#
The distance of P from #(3,0)#
#d_2=sqrt((h-3)^2+(k-0)^2)#
By the given condition
#d_1+d_2=8#
#=>sqrt((h+3)^2+(k-0)^2)+sqrt((h-3)^2+(k-0)^2)=8#
#=>sqrt((h+3)^2+k^2)=8-sqrt((h-3)^2+k^2)#
Squaring both sides we have
#=>(h+3)^2+k^2=64-16sqrt((h-3)^2-k^2)+(h-3)^2+k^2#
#=>(h+3)^2+k^2-(h-3)^2-k^2=64-16sqrt((h-3)^2+k^2)#
#=>(h+3)^2-(h-3)^2=64-16sqrt((h-3)^2+k^2)#
#=>12h=64-16sqrt((h-3)^2+k^2)#
#=3h=16-4sqrt((h-3)^2+k^2)#
#=>4sqrt((h-3)^2+k^2)=16-3h#
#=>16((h-3)^2+k^2)=(16-3h)^2#
#=>16((h^2-6h+9)+k^2)=256-96h+9h^2#
#=>16h^2-96h+144+16k^2=256-96h+9h^2#
#=>7h^2+16k^2=112#
Dividing both sides by 112
#=>h^2/16+k^2/7=1#
Replacing 'h' by 'x' and 'k' by ' y' we have the equation of the locus of point P
#=>x^2/16+y^2/7=1 " which is equation of an ellipse "#
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Question No 2
The given equations of straight lines are
#ax+by+c=0 and (a-b)x+(a+b)y+c=0#
Transforming these equations into #y=mx +c " forms"# we have
#y=-a/bx-c/b.....(1)->"slope " m_1=-a/b#
#y=-(a-b)/(a+b)x-c/(a+b)....(2)->"slope " m_2=-c/(a+b)#
Angle between the line (1) and (2)
#theta =tan^(-1)((m_2-m_1)/(1+m_1m_2))#
#=tan^(-1)((-c/(a+b)+a/b)/(1+(ac)/(b(a+b))))#
#=tan^(-1)((a^2+ab-bc)/(b^2+ab+ac))#
