Question

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

Options

• 22/7

• 6/7

• `-6`

• none of these

Answer 1

6/7

 

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

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

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

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

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

\[\text { Now }, \]

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

\[\text { Let us solve one of these to get the value of }t.\]

\[ t^2 + 3t - 10 = 0 \text { and } 2 t^2 - 2t - 4 = 0\]

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

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

\[\text { These two have t = 2 as a common solution } . \]

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