Question

slope of tangent ( =frac{d y}{d x}=frac{d y / d t}{d x / d t} )
( frac{d y}{d t}=4 t+3, frac{d x}{d t}=2 t-3 )
( left.frac{d y}{d x}right|_{(-1,10)}=y cdot frac{4++3}{2 t-3} )
( y=1 quad y=-1 Rightarrow quad t^{2}-3 t+1=-1 )
( Rightarrow quad quad Rightarrow quad t^{2}-3 t+2=0 )
[
begin{array}{c}
(t-1)(t-2)=0
t=1,2
2 t^{2}+3 t-4=10
2 t^{2}+3 t-14=0
(t-2)(2 t+7)
end{array}
]
( t=2, t=-7 / 2 )
now putting ( t=2 quad ) in ( frac{d y}{d x} )
[
left.frac{d y}{d n}right|_{(-1,10)}=frac{p+3}{4-3}=11
]

# The slope of tangent to the curve represented parametrically by the equations x = 2 - 3t + 1 and y= 22 + 3t - 4 at M(-1,10) is - (A) 5 (B) 9 (C)7 (D) 11

Solution