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Question

The normal at any point P(x,y) of a curve meets the x-axis at Q and N is the foot of the ordinate at P.If NQ=x(1+y2)1+x2, then equation of such curve, given that it passes through the point (3,1) is:

A
x2y2=8
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B
x2+2y2=11
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C
x25y2=4
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D
x2+3y2=7
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Solution

The correct option is C x2−5y2=4The equation of the normal at P, can be written as Y−y=−1(dy/dx)(X−x)The normal to the curve meets the x-axis at Q(x+ydydx,0). N is the foot of the ordinate. Hence, N has the coordinates (x,0).NQ=ydydx=x(1+y2)1+x2∫y1+y2dy=∫x(1+y2)1+x2dx12ℓn(1+y2)=12ℓn(1+x2)+12ℓnc(1+y2)=c(1+x2)The given curve passes through (3,1). Hence, 2=c(10)⇒c=15∴5(1+y2)=1+x2x2−5y2=4

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