A curve has the parametric equations $x=2t^2$ and $y=4t$. What is the value(s) of $k$ such that $y=x+k$ is a tangent to the curve?

Question

A curve has the parametric equations $x=2t^2$ and $y=4t$. Find the value(s) of $k$ such that $y=x+k$ is a tangent to the curve.

I get that you need to use differentiation to do this and I've tried doing these before with questions such as this one: Finding the values of k for an equation which is a tangent to a curve but I'm not exactly sure how to figure this one out. Please help ASAP and please be detailed in the working out so I can understand :)

Thank you!

Answer 1

I assume you know parametric differentiation using chain rule.

The requirement $y=x+k$ is tangent to the curve essentially means the slope, $\frac{dy}{dx}$, of both curves are equal. The slope of the line is $1$, and slope of curve at $(2t^2,4t)$ is:

$$\frac{dy}{dx}=\frac{dy}{dt} \cdot \frac{1}{\frac{dx}{dt}} = \fbox{Try it yourself} =^{\textrm{Want to equal to}} 1$$

From that you get the value of $t$, and thus the point $(2t^2,4t)$, which you can then substitute into the line equation $y=x+k$ to get the value of $k$.

Cheers! :)

Answer 2

First, we have to pass from the parametric form, to linear form. In particular, we can substitute $t=\frac{y}{4}$, so: $$x=\frac{y^2}{8}$$ This is a parabola, with vertex in $O(0,0)$ and so the tangent lines are given by: $$x=\frac{(x+k)^2}{8}$$ with $\Delta=0$, or: $$4(k-4)^2-4k^2=0\leftrightarrow 4k^2-32k+64-4k^2=0\leftrightarrow 32k=64\leftrightarrow k=2$$ The only line is so: $$y=x+2$$

Answer 3

It seems you're probably having difficulty with the differentiation so let me start you off with a hint.

Using chain rule (this is the basis of parametric differentiation), $\frac {dy} {dx} = \frac {\frac{dy} {dt}} {\frac {dx} {dt}} $.

First start by working that out in terms of the parameter $t$.

Equate it to the slope of the tangent ($1$). What is $t$ at this point?

Can you finish?

Answer 4

Note that the tangent slope is 1, which leads to

$$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac1t=1\implies t=1$$

Then, the tangent point is $(2t^2,4t)=(2, 4)$. Plug into $y=x+k$ to obtain $k=2$.