The slope of the tangent to the curve $x = t^{2} + 3 t - 8, y = 2 t^{2} - 2 t - 5$ at the point t = 2 is

\frac{7}{6}

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$\frac{5}{6}$

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$\frac{6}{7}$

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$1$

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Solution

The correct option is C
$\frac{6}{7}$

We have,
$\frac{d x}{d t} = 2 t + 3 a n d \frac{d y}{d t} = 4 t - 2 \frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{4 t - 2}{2 t + 3}$
Thus, slope of the tangent to the curve at the point t = 2 is
${[\frac{d y}{d x}]}_{t - 2} = \frac{4 (2) - 2}{2 (2) + 3} = \frac{6}{7}$
Thus, slope of the tangent to the curve at the point t = 2 is
Hence (c) is the correct answer

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Similar questions

Q. The slope of tangent to the curve

x = t^{2} + 3 t - 8, y = 2 t^{2} - 2 t - 5

at the point

(2, - 1)

is :

The slope of the tangent to the curve $x = t^{2} + 3 t - 8, y = 2 t^{2} - 2 t - 5$ at the point t = 2 is

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

(a) $\frac{22}{7}$
(b) $\frac{6}{7}$
(c) $\frac{7}{6}$
(d) $- \frac{6}{7}$

Q. Find the slope of the tangent to the curve x = t² + 3t − 8, y = 2t² − 2t − 5 at t = 2.

Q. What is the slope of the tangent to the curve

x = t^{2} + 3 t - 8, y = 2 t^{2} - 2 t - 5

at t = 2 ?

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

The correct option is C 67 We have, dxdt=2t+3 and dydt=4t−2dydx=dy/dtdx/dt=4t−22t+3 Thus, slope of the tangent to the curve at the point t = 2 is [dydx]t−2=4(2)−22(2)+3=67 Thus, slope of the tangent to the curve at the point t = 2 is Hence (c) is the correct answer

The slope of the tangent to the curve $x = t^{2} + 3 t - 8, y = 2 t^{2} - 2 t - 5$ at the point t = 2 is