The slope of the tangent to the curve x = t2 + 3t − 8, y = 2t2 − 2t − 5 at the point (2, −1) is _____________ .

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$\frac{6}{7}$ Given: x = t2 + 3t − 8 and y = 2t2 − 2t − 5 \Rightarrow dx/dt = 2t + 3 \text { and } dy/dt = 4t - 2 \therefore dy/dx = dy/dt/dx/dt = 4t - $\frac{2}{2}$t + 3 \text { The given point is } (2, -1). \therefore x = 2 \text { and } y = -1 \text { Now }, t^2 + 3t - 8 = 2 \text { and }2 t^2 - 2t - 5 = - 1 \Rightarrow t^2 + 3t - 10 = 0 \text { and } t^2 - t - 2 = 0 \Rightarrow \left( t + 5 \right)\left( t - 2 \right) = 0 \text { and } \left( t + 1 \right)\left( t - 2 \right) = 0 \Rightarrow t = - 5, 2 \text { and }t = - 1, 2 \text { We can see that t = 2 satisfies both of these }. \therefore \text { Slope of the tangent } = \left( dy/dx \right)_{t = 2} =8 - $\frac{2}{4}$ + 3=$\frac{6}{7}$

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