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) 7/6  (D) - 6/7

Solution:

The given curve is

x = t² + 3t - 8 and y = 2t² - 2t - 5

Therefore,

On differentiating both the functions with respect to t, we get

dx / dt = 2t + 3

dy / dt = 4t - 2

Hence,

dy/dx = (dy/dt) / (dx/dt)

= (4t - 2) / (2t + 3)

The given point is (2, - 1)

At x = 2 , we have:

t² + 3t - 8 = 2

⇒ t² + 3t - 10 = 0

⇒ (t - 2)(t + 5) = 0

⇒ t = 2 or t = - 5

At y = -1 ,we have:

2t ² - 2t - 5 = - 1

⇒ 2t ² - 2t - 4 = 0

⇒ 2 (t ² - t - 2) = 0

⇒ (t - 2)(t + 1) = 0

⇒ t = 2 or t = - 1

The common value is t = 2

Hence,

the slope of the tangent to the given curve at point (2, - 1) is

dy/dx]_{t = 2} = (4(2) - 2)/(2(2) + 3)

= (8 - 2) / (4 + 3)

= 6 / 7

Thus, the correct option is B

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise ME Question 20

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) 7/6 (D) - 6/7

Summary:

The slope of the tangent to the curve x = t² + 3t - 8, y = 2t² - 2t - 5 at the point  (2, - 1) is 6/7.  Thus, the correct option is B