# Find an equation of the tangent to the curve at the point corresponding to the given parameter: `x=t-t^(-1)` , `y=5+t^2` , `t=1`

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Expert Answers

justaguide | Certified Educator

The curve is defined by `x = t - t^(-1)` and `y = 5 + t^2`

`dx/dt = 1 + 1/t^2`

`dy/dt = 2t`

`dy/dx = (dy/dt)/(dx/dt) = (2t)/(1 + 1/t^2)`

At the point where t = 1, `dy/dx = 2/2 = 1`

The slope of the graph with respect to the value of t varies as shown in the following graph:

The tangent to the curve at the point where t = 1 has a slope of 1. The equation of the tangent is `(y - 6)/(x - 0) = 1`

=> y - 6 = x

=> x - y + 6 = 0

**The required equation of the tangent is x - y + 6 = 0**