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Find the equation of tangent to the curv...

Find the equation of tangent to the curve by x = `t^(2) + t + 1 , y = t^(2) - t + 1` at a point where t = 1

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Slope of tangent = `(dy)/(dx)`
Here `x = t^(2) + t + 1`
`implies (dx)/(dt) = (2t +1)`
`implies y = t^(2) - t + 1`
So , `(dy)/(dx) = ((2t-1)/(2t+1))`
`implies (dy)/(dx)|_(t=1) = (1)/(3)`
Equation of tangent will be given by
`(y-1) = (1)/(3) (x-3)`
`implies 3y - x = 0`
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