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Question

Transform the equation xa+yb=1 into normal form when a>0 and b>0. If the perpendicular distance of stright line form the origin is p, deduce that 1p2=1a2+1b2

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Solution

xa+yb=1.......(i),a>0, b>0
1a2+1b2=a2+b2a2b2=a2+b2ab
Divide (i) throughout by a2+b2ab
xaa2+b2ab+yba2+b2ab=1a2+b2ab
xba2+b2+yaa2+b2=aba2+b2...........(ii)
Slope of (ii) is given by
tanα=Coefficient of xCoefficient of y
=1a1b=ba which is negative, hence α is obtuse
tanα=ba=PerpendicularBase
Hypotenuse=Base2+Perpendicular2=a2+b2
For obtuse angles sin is positive and cos is negative
sinα=ba2+b2
cosα=ba2+b2
Hence (ii) will become
x(cosα)+y(sinα)=aba2+b2
We know that
sin(A)=sin(πA)
and cos(A)=cos(πA)
xcos(πα)+ysin(πα)=aba2+b2
Which is the required normal form where tanα is slope of given line
Hence right hand side must give perpendicular distance of the straight line from the orogin.
P=aba2+b2
P2=a2b2a2+b2
1P2=a2+b2a2b2
1P2=a2a2b2+b2a2b2
1P2=1a2+1b2

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