Question

Find the equations of the tangents to the curve x = 9t2 + 6, y = 6t3 + 3 that pass through the point (15, 9).

Answer 1

Tangents to the curve have slope $\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}$. We have

$\dfrac{\mathrm dy}{\mathrm dt}=18t^2$

$\dfrac{\mathrm dx}{\mathrm dt}=18t$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{18t^2}{18t}=t$

The point (15, 9) occurs on the curve and corresponds to

$\begin{cases}9t^2+6=15\\6t^3+3=9\end{cases}\implies\begin{cases}t=\pm1\\t=1\end{cases}\implies t=1$

so that the slope of the line tangent to the curve at (15, 9) is 1. Then the equation of this tangent line is

$y-9=1(x-15)\implies y=x-6$