Answer
The equations of the tangent lines are:
$y = x-1$
$y = -2x+11$
Work Step by Step
$x = 3t^2+1$
$y = 2t^3+1$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{6t^2}{6t} = t$
The equation of the tangent line would have this form:
$y-3 = t(x-4)$
$2t^3+1-3 = t(3t^2+1-4)$
$2t^3-2 = 3t^3-3t$
$t^3-3t+2 = 0$
$(t-1)(t^2+t-2) = 0$
$(t-1)(t+2)(t-1) = 0$
$t = 1, -2$
When $t = 1$, then $\frac{dy}{dx} = t = 1$
$x = 3(1)^2+1 = 4$
$y = 2(1)^3+1 = 3$
We can find the equation of the tangent line when $t=1$:
$y-3 = (1)(x-4)$
$y = x-1$
When $t = -2$, then $\frac{dy}{dx} = t = -2$
$x = 3(-2)^2+1 = 13$
$y = 2(-2)^3+1 = -15$
We can find the equation of the tangent line when $t = -2$:
$y-(-15) = -2(x-13)$
$y+15 = -2x+26$
$y = -2x+11$
The equations of the tangent lines are:
$y = x-1$
$y = -2x+11$
