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y=x^3-7x=6 at its point of intersection with the x-axis
x^2+y^2-6x+2y=0;(0,0)
sol- 1
let the tangent be at (h,0) then, it satisfies the given curve;
h^3-7h -6=0 ......................................... (i)
& dy/dx = 3x^2-7
=3h^2-7
now, eqn of line having slope (3h^2-7) & passing through (h,0) is
(3h^2-7)x - y=3h^3-7h ................................ (ii)
from (i) we have h^3-7h-6=0
or (h+1)(h+2)(h-3)=0
or h= -1,-2, 3 putting it in eqn (ii) we get three tangents
4x+y-10=0
5x-y+31=0 &
20x-y -74=0
(there are three tangents as the curve cuts the x-axis at three points (-2,0), (-1,0), (3,0))
sol - (ii)
given eqn represents a circle
so, eqn of tangent at (a,b) is given by ((a,b) lies on the curve)
xa + yb - 6(x+a)/2 +2(y+b)/2=0 put a=0 b=0 , given
3x-y=0
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