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प्रश्न
The slope of the tangent to the curve x = t2 + 3 t − 8, y = 2t2 − 2t − 5 at point (2, −1) is ________________ .
पर्याय
22/7
6/7
`-6`
none of these
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उत्तर
6/7
\[x = t^2 + 3t - 8 \text { and } y = 2 t^2 - 2t - 5\]
\[\frac{dx}{dt} = 2t + 3 \text { and } \frac{dy}{dt} = 4t - 2\]
\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4t - 2}{2t + 3}\]
\[\text { The given point is } (2, -1).\]
\[\therefore x=2 \text { and }y=-1\]
\[\text { Now }, \]
\[ t^2 + 3t - 8 = 2 \text { and }2 t^2 - 2t - 5 = - 1\]
\[\text { Let us solve one of these to get the value of }t.\]
\[ t^2 + 3t - 10 = 0 \text { and } 2 t^2 - 2t - 4 = 0\]
\[ \Rightarrow \left( t + 5 \right)\left( t - 2 \right) = 0 \text { and } \left( 2t + 2 \right)\left( t - 2 \right) = 0\]
\[ \Rightarrow t = - 5 \ or \ t=2 \text { and }t=-1 \ or \ t=2\]
\[\text { These two have t = 2 as a common solution } . \]
\[ \therefore \text { Slope of the tangent } = \left( \frac{dy}{dx} \right)_{t = 2} = \frac{8 - 2}{4 + 3} = \frac{6}{7}\]
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