A curve is given parametrically by x = t² + 3*t - 8 and y = 2*t² - 2*t - 5. The slope of the tangent to the curve at the point M(2, -1) equals lambda / 7. Find lambda.

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6. At point M(2,-1): solve t²+3t-8 = 2 => t²+3t-10 = 0 => (t-2)(t+5)=0 => t=2 (since t=-5 gives y=55-10-5=40 != -1). At t=2: dy/dt = 4(2)-2 = 6; dx/dt = 2(2)+3 = 7. Slope = dy/dx = 6/7. Given slope = lambda/7, so lambda = 6.

JEE Advanced Maths: Differential Calculus / Applications of Derivatives questions with solutions

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