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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)

23 Points
9 years ago

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