20) The slope of the tangent to the curve x = t^2 + 3t - 8, y = 2t^2 - 2t - 5 at the point
(2,– 1) is

 A ) 22/7 

B ) 6/7 

C ) 7/6 

D ) -6 /7 

Answers (1)

Given curves are 
x = t^2 + 3t - 8 \ and \ y = 2t^2 - 2t - 5
At point (2,-1)
t^2 + 3t - 8 = 2\\ t^2+3t-10=0\\ t^2+5t-2t-10=0\\ (t+5)(t-2) = 0\\ t = 2 \ and \ t = 5
Similarly,
2t^2-2t-5 = -1\\ 2t^2-2t-4=0\\ 2t^2-4t+2t-4=0\\ (2t+2)(t-2)=0\\ t = -1 \ and \ t = 2
The common value between two is t = 2 
Hence, we find the slope of the tangent at t = 2
We know that the slope of the tangent at a  given point is given by \frac{dy}{dx}
\frac{dy}{dt} = 4t - 2
\frac{dx}{dt} = 2t + 3
\left ( \frac{dy}{dx} \right )_{t=2} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4t-2}{2t+3} = \frac{8-2}{4+3} = \frac{6}{7}
Hence, (B) is the correct answer

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