Question

How do you find the equations of the tangents to that curve `x=3t^2 + 1` and `y=2t^3 + 1` that pass through point (4,3)?

Answer 1

Tangent Line
`y=x-1`

Explanation:

We find the equation first consisting only of x and y
by eliminating variable t.

Given `x=3t^2+1" "`first equation
and `y=2t^3+1" "`second equation

Use the first equation then substitute its equivalent in the second equation

`x=3t^2+1" "`first equation
`t=((x-1)/3)^(1/2)" "`first equation

`y=2t^3+1" "`second equation
`y=2(((x-1)/3)^(1/2))^3+1" "`second equation

We now have y in terms of x
`y=2((x-1)/3)^(3/2)+1" "`

Solve for the slope `m=dy/dx`

`m=dy/dx=2*3/2*((x-1)/3)^(3/2-1)*d/dx((x-1)/3)+d/dx(1)`

`m=dy/dx=cancel2*cancel3/cancel2*((x-1)/3)^(1/2)*(1/cancel3)+0`

`m=((x-1)/3)^(1/2)" "`at `(4, 3)`

`m=((4-1)/3)^(1/2)" "`

`m=1`

The Tangent Line using `(x_1, y_1)=(4, 3)` and `m=1`

`y-y_1=m(x-x_1)`

`y-3=1(x-4)`

`y=x-4+3`

`y=x-1`

Kindly see the graphs of the curve `y=2((x-1)/3)^(3/2)+1" "`
and the tangent line `y=x-1`.
graph{(y-2((x-1)/3)^(3/2)-1)(y-x+1)=0[-10, 10, -5,5]}

God bless...I hope the explanation is useful.