Tangents with Parametric Equations — Question 10

Question 10 – Solution

Solution

1. To meet the $x$-axis at $t=1$, the $y$-coordinate must be zero: $$y(1)=1^3-a(1)=1-a=0.$$ Therefore, the position condition requires $$a=1.$$

2. Now impose the slope condition. Differentiate: $$\frac{dx}{dt}=1, \qquad \frac{dy}{dt}=3t^2-a.$$ Since $dx/dt=1$, the tangent slope is $$\frac{dy}{dx}=3t^2-a.$$

3. At $t=1$, requiring slope $2$ gives $$3-a=2 \quad\Longrightarrow\quad a=1.$$ Thus the position and slope conditions agree.

4. Verify the result directly. With $a=1$, $$y(1)=1-1=0, \qquad \left.\frac{dy}{dx}\right|_{t=1}=3-1=2.$$ The point is $(x(1),y(1))=(1,0)$, and its tangent line is $y=2(x-1)$.

5. Hence the design is possible, and the unique value is $$\boxed{a=1}.$$