Tangents with Parametric Equations — Question 9

Question 9 – Solution

Solution

1. First find every parameter value that could produce the point $(1,0)$. From $x=t^2=1$, $$t=1\quad\text{or}\quad t=-1.$$

2. Check the $y$-coordinate for both values: $$y(1)=1^4-1=0, \qquad y(-1)=(-1)^4-(-1)=2.$$ Only $t=1$ produces $(1,0)$, so there can be only one tangent line there.

3. Differentiate: $$\frac{dx}{dt}=2t, \qquad \frac{dy}{dt}=4t^3-1.$$ At $t=1$, $dx/dt=2\ne0$, and the slope is $$\frac{dy}{dx} =\frac{4t^3-1}{2t} =\frac{4(1)^3-1}{2(1)} =\frac32.$$

4. Apply point–slope form through $(1,0)$: $$y-0=\frac32(x-1).$$ Therefore, the only tangent line at the specified point is $$\boxed{y=\frac32(x-1)}.$$