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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?

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3 Answers 3

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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 $t=1,-2$.

enter image description here

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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.

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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

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  • $\begingroup$ So, would you say that $y=x^3$ has no tangent at the origin? $\endgroup$ Commented Jun 4, 2018 at 7:22
  • $\begingroup$ You appear to be making the same error that the O.P. did: the original problem doesn’t say that $(4,3)$ is the point of tangency, only that the tangent lines must pass through this point. $\endgroup$ Commented Jun 4, 2018 at 7:31
  • $\begingroup$ Right, through (4,3) there is tangent at that point but slope form of line $(y-3)=-2(x-4)$ equation crosses curve not touching it. Also there is difference between secant ,tangent line and normal line definition. $(y-3)=-2(x-4)$ This line doesn't fit the definition of tangent line as you can see in graph. Im not so quiet sure about the origin you mentioned $y=x^3$ but i guess it will not be called as tangent if we take $x=0 and y=0$ line equation. tutorial.math.lamar.edu/Classes/CalcI/Tangents_Rates.aspx $\endgroup$ Commented Jun 4, 2018 at 7:57
  • $\begingroup$ You’re misinterpreting what that page you’ve cited is telling you. That line in the illustration is certainly not tangent to the curve at the point that it crosses the curve but it is tangent to the curve elsewhere. The same situation exists in this problem. The second line is not tangent to the curve at $(4,3)$, but at some other point. The fact that it crosses the curve somewhere else is immaterial. In particular, the idea that a tangent line doesn’t cross the curve at the point of tangency only holds for locally convex curves, which is not the case for my earlier example. $\endgroup$ Commented Jun 4, 2018 at 17:46
  • $\begingroup$ I guess you are jumbling between touching and crossing,of course the second line is not tangent as you can clearly see it was crossing given curve. But the first line is touching exactly at (4,3). The point here im making is that for a given point there only exist one tangent line not the multiple tangent line at common point. More likely first line is a Secant Line as if we extended , but as in problem we derived tangent line so first line is the only line which will fit in the definition of tangent line $\endgroup$ Commented Jun 4, 2018 at 17:56

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