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Question

Let P(x0,y0) be a point on the curve C:(x211)(y+1)+4=0, where x0,y0N. If area of the triangle formed by the normal drawn to the curve C at P and the co-ordinate axes is ab, where a,bN, then the least value of a6b is

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Solution

(x211)(y+1)+4=0
(x211)(y+1)=4=2×2
On comparing, we get
P(x0,y0)(3,1) as x0,y0N
y=8x(x211)2
yx=3=6
Slope of normal, mN=16

Equation of normal at P(3,1) is
y1=16(x3)
x+6y=9


Area =12×9×32=274

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