Grade 11Analytical Geometry
The equation of the tangent to the parabola y=x^2-x at the point where x=1 is?
1 Answer
Arun
Dear Piyush
At x = 1
y = 1² -1 = 1- 1 = 0
Hence we have to find a tangent at (1,0)
Now
Equation of tangent -
(y + 0)/2 = x * 1 - (x+1)/2
y/2 = (2x - x -1)/2
y = x -1
This is the required tangent
Regards
Arun (askIITians forum expert)
At x = 1
y = 1² -1 = 1- 1 = 0
Hence we have to find a tangent at (1,0)
Now
Equation of tangent -
(y + 0)/2 = x * 1 - (x+1)/2
y/2 = (2x - x -1)/2
y = x -1
This is the required tangent
Regards
Arun (askIITians forum expert)
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