Find equations of the tangents to the curve $x=3t^2+1,\space y=2t^3+1$ that pass through the point $(4,3)$
My attempt:
If we have $x=3t^2+1,\space y=2t^3+1$, then \begin{cases} 4=3t^2+1\\3=2t^3+1 \end{cases}$\implies$\begin{cases} t=1,-1\\ t=1 \end{cases}Therefore we take the parameter of intersection for point $(4,3)$ which is $t=1.$ $$Recall \space \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{6t^2}{6t}=t$$
So at point $(4,3)$ the slope of the tangent line is $\frac{dy}{dx}=t=1$.
Thus using point slope form: $$y-y_0=\frac{dy}{dx}\big(x-x_0\big)\implies
y-3=x-4\implies y=x-1$$
Our tangent line then is $y=x-1$. However, it can be shown through the elimination of $x$ and $y$ from the tangent line equation that $y=-2x+11$ is also tangent to the curve and passes through $(4,3)$.
My question is:
How do we know a priori if there exists multiple tangents to a curve through a common point? If it weren't for a solutions manual, I would not think to eliminate $x$ and $y$ to solve for the slope of the tangent. What is the reason there are multiple tangents through a common point? Is this specific to parametric curves?
Best Answer
We can try to solve for it.
$$x=3t^2+1$$ $$y=2t^3+1$$
We have
$$\frac{dy}{dx}=t$$
Hence the tangent line that we are looking for has the form of
$$(y-3)=t(x-4)$$
$$(2t^3+1-3)=t(3t^2+1-4)$$ Note that this is a cubic equation, hence we should expect up to $3$ solutions for $t$.
$$(2t^3-2)=t(3t^2-3)$$
$$2(t^3-1)=3t(t^2-1)$$
$$2(t-1)(t^2+t+1)=3t(t-1)(t+1)$$
$$(t-1)(2t^2+2t+2)=(t-1)(3t^2+3t)$$
$$(t-1)(t^2+t-2)=0$$
$$(t-1)(t+2)(t-1)=0$$
$$(t-1)^2(t+2)=0$$
Hence, we would check when $t=1,-2$.