Answer:
The equation of the straight line
x - 24y + 38 =0
Step-by-step explanation:
Step(i):-
Given that x(t) = t²+1 ..(i)
and y(t) = √1+t ..(ii)
Differentiating equation(i) with respective to 't'
[tex]\frac{dx}{dt} = 2t[/tex]
Differentiating equation(ii) with respective to 't'
[tex]\frac{dy}{dt} = \frac{1}{2\sqrt{1+t} }[/tex]
Step(ii):-
The slope of the tangent
[tex]m = \frac{dy}{dx} = \frac{\frac{dy}{dt} }{\frac{dx}{dt} } = \frac{\frac{1}{2\sqrt{1+t} } }{2t}[/tex]
[tex]m =( \frac{dy}{dx} )_{t=3} = \frac{1}{4(3)\sqrt{1+3} }[/tex]
[tex]m = \frac{1}{24}[/tex]
Step(iii):-
Point x = t²+1 = 3²+1 = 10
y = √1+t =√1+3 = √4 =2
The point on the tangent line is ( 10 ,2)
The equation of the straight line
[tex]y - y_{1} = m( x-x_{1} )[/tex]
[tex]y - 2 = \frac{1}{24} ( x-10 )[/tex]
24 (y-2) = x-10
24y - 48 = x-10
x - 24 y -10 +48 =0
x - 24y + 38 =0
Final answer:-
The equation of the straight line
x - 24y + 38 =0
