Question

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

Answer 1

HELLO DEAR,

given curve is x = t² + 3t - 8 , y = 2t² - 2t - 5.

now, dx/dt = 2t + 3 , dy/dt = 4t - 2

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

now the points are (2 , -1)
at x = 2

t² + 3t - 8 = 2

t² + 3t - 10 = 0

t² + 5t - 2t - 10 = 0

t(t + 5) - 2(t + 5) = 0

(t - 2)(t + 5) = 0

t = 2 , t = -5
now, y = -1

-1 = 2t² - 2t - 5

2t² - 2t - 4 = 0

t² - t - 2 = 0

t² - 2t + t - 2 = 0

t(t - 2) + 1(t - 2) = 0

(t + 1)(t - 2) = 0

t = 2 , t = -1

The common value of t is 2. 
Hence, the slope of the tangent to the given curve at point (2, −1)

$\bold{\frac{dy}{dx}_{x = 2} = \frac{4(2) - 2}{2(2) + 3}}$

$\bold{\frac{8 - 2}{4 + 3} = \frac{6}{7}}$

$\bold{\large{\boxed{HENCE,\;\; OPTION\;\;(B)\;\;IS\;\; CORRECT}}}$

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer:

Step-by-step explanation:

Hope you understood