Find equations of the tangents to a parametric curve that pass through a given point

Question

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?

Answer 1

It’s entirely normal to have multiple tangents through a given point, as opposed to multiple tangents at a point on a curve. For example, there are two tangents to a circle through any point exterior to that circle.

Having the curve given in parametric form isn’t really the issue here. What’s happened is that you’ve solved a somewhat different problem than the one that was posed: you computed the tangent at the point $(4,3)$, but the task was to compute the tangents that pass through this point. This is conceptually no different than, say, finding the tangents to $y=x^3-4x$ that pass through $(2,0)$: there is a single tangent at that point, but there’s also a different line through $(2,0)$ that’s tangent to the curve at $(-1,3)$. That the point happens to be on the curve is either coincidental or a deliberate choice on the part of whoever wrote this problem to trap the unwary. By looking only at the tangent at the point you discarded other potential solutions right from the start.

Answer 2

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 $t=1,-2$.

Answer 3

Tangent line definition is a straight line or plane that touches a curve or curved surface at a point, but if extended does not cross it at that point. So here even though $(y-3)=-2(x-4)$ satisfies line equation but since it crosses curve at $(4,3)$ its not a tangent line. There are no multiple tangent lines to the curve which passes through common point