Question

1. find the points on the graph of $y^{3}-27y = x^{2}-90$ where the tangent line is vertical.

Explanation:

Step1: Differentiate implicitly

Differentiate $y^{3}-27y = x^{2}-90$ with respect to $x$.
Using the chain - rule, we have $3y^{2}y'-27y'=2x$.
Factor out $y'$: $y'(3y^{2}-27)=2x$.
So, $y'=\frac{2x}{3y^{2}-27}$.

Step2: Find when the tangent is vertical

A tangent line is vertical when the slope is undefined, i.e., when the denominator of $y'$ is zero and the numerator is non - zero.
Set $3y^{2}-27 = 0$.
Solve for $y$:
\[

$$\begin{align*} 3y^{2}-27&=0\\ 3y^{2}&=27\\ y^{2}&=9\\ y&=\pm3 \end{align*}$$

\]
When $y = 3$:
Substitute $y = 3$ into the original equation $y^{3}-27y=x^{2}-90$.
$3^{3}-27\times3=x^{2}-90$.
$27 - 81=x^{2}-90$.
$- 54=x^{2}-90$.
$x^{2}=36$, so $x=\pm6$.
When $y=-3$:
Substitute $y = - 3$ into the original equation $y^{3}-27y=x^{2}-90$.
$(-3)^{3}-27\times(-3)=x^{2}-90$.
$-27 + 81=x^{2}-90$.
$54=x^{2}-90$.
$x^{2}=144$, so $x=\pm12$.

Answer:

The points are $(6,3),(-6,3),(12, - 3),(-12,-3)$