Answer
$y=x-1$ and
$y=-2x+11$
are tangents to the curve that pass through (4,3)
Work Step by Step
$x=3t^{2}+1$, $\displaystyle \frac{dx}{dt}=6t$,
$y=2t^{3}+1,\ \displaystyle \frac{dy}{dt}=6t^{2}$
$\displaystyle \frac{dy}{dx}=\frac{6t^{2}}{6t}=t$.
At point ($x_{1},y_{1}$), the tangent line has equation
$y-y_{1} =t(x-x_{1})$
... representing $x_{1}$ and $y_{1}$ with t:
$y-(2t^{3}+1)=t[x-(3t^{2}+1)]$.
A tangent line passing through the point (4,3) satisfies
$3-(2t^{3}+1)=t[4-(3t^{2}+1)]$
$-2t^{3}+2=4t-3t^{3}-t$
$t^{3}-3t+2=0$
$t^{3}-t-2t+2=0$
$t(t^{2}-1)-2(t-1)=0$
$(t-1)(t(t+1)-2)=0$
$(t-1)(t^{2}+t-2)=0$
$(t-1)(t-1)(t+2)=0$
$(t-1)^{2}(t+2)=0$
$t=1$ or $t=-2$.
When $t=1$ the point is $(3(1^{2})+1,2(1^{3})+1)= (4,3)$
So one equation is the tangent to the curve at (4,3):
$y-(2\cdot 1+1)=(1)[x-(3(1)+1)]$.
$y-3=x-4$,
$y=x-1$
When $t=-2$ the point is $(3(-2)^{2}+1, 2(-2)^{3}+1)=(-11,-15)$
So, another is the tangent in the point $(-11,-15)$:
$y-(2\cdot(-8)+1)=(-2)[x-(3(4)+1)]$.
$y-15=-2x+26$
$y=-2x+11$
