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Question: The slope of the tangent to the curve x = t<sup>2</sup> + 3t – 8, y = 2t<sup>2</sup>– 2t – 5 at the...

The slope of the tangent to the curve

x = t2 + 3t – 8, y = 2t2– 2t – 5 at the point (2, – 1) is –

A

22/7

B

6/7

C

– 6

D

None

Answer

6/7

Explanation

Solution

dydx=dy/dtdx/dt=4t22t+3\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t - 2}{2t + 3}

at x = 2 Ž 2 = t2 + 3t – 8 Ž t2 + 3t – 10 = 0

= –5, 2

at y = –1 Ž –1 = 2t2 – 2t – 5 Ž 2t2 – 2t – 4 = 0

Ž t = 2, – 1

\ put t = 2

\ dydx=67\frac{dy}{dx} = \frac{6}{7}