Question

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

Options

• \[\frac{22}{7}\]

• \[\frac{6}{7}\]

• \[\frac{7}{6}\]

• \[- \frac{6}{7}\]

Answer 1

\[\frac{6}{7}\]

 

Given:
x = t² + 3t − 8 and y = 2t² − 2t − 5

\[\Rightarrow \frac{dx}{dt} = 2t + 3 \text { and } \frac{dy}{dt} = 4t - 2\]

\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4t - 2}{2t + 3}\]

\[\text { The given point is } (2, -1).\]

\[\therefore x = 2 \text { and } y = -1\]

\[\text { Now }, \]

\[ t^2 + 3t - 8 = 2 \text { and }2 t^2 - 2t - 5 = - 1\]

\[ \Rightarrow t^2 + 3t - 10 = 0 \text { and } t^2 - t - 2 = 0\]

\[ \Rightarrow \left( t + 5 \right)\left( t - 2 \right) = 0 \text { and } \left( t + 1 \right)\left( t - 2 \right) = 0\]

\[ \Rightarrow t = - 5, 2 \text { and }t = - 1, 2\]

\[\text { We can see that t = 2 satisfies both of these }.\]

\[ \therefore \text { Slope of the tangent } = \left( \frac{dy}{dx} \right)_{t = 2} =\frac{8 - 2}{4 + 3}=\frac{6}{7}\]