find the equation of tangent and normal of the curve y=X^ +2x-1 at (1,2)
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Answered by
1
Answer:
The given equation of curve is,
y=x2
Thus, slope of tangent to the curve is given by,
m=dxdy
∴m=dxd(x2)
∴m=2x
Slope of tangent at point (1,1) is,
m=2×1
∴m=2
Thus, equation of tangent at point (1,1) is given by two point form as,
y−y1=m(x−x1)
y−1=2(x−1)
∴y−1=2x−2
∴2x−y−1=0
∴y=2x−1
Answered by
1
Answer:
Equation of tangent=y-2=4(x-1)
=y-4x+2=0
Equation of normal=y-2=-1/4(x-1)
=4y+x=9
Step-by-step explanation:
y=x^2+2x-1
slope of curve (dy/dx) at (1,2)=2x+2
=4
Equation of tangent=y-2=4(x-1)
=y-4x+2=0
Equation of normal=y-2=-1/4(x-1)
=4y+x=9
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