Answer
The points where the tangent line is vertical are $(1,0)$, $(-1,0)$, $\left(\dfrac{\sqrt{3}}{2},\dfrac{1}{\sqrt2}\right)$, $\left(-\dfrac{\sqrt{3}}{2},\dfrac{1}{\sqrt2}\right)$, $\left(\dfrac{\sqrt{3}}{2},-\dfrac{1}{\sqrt2}\right)$ and $\left(-\dfrac{\sqrt{3}}{2},-\dfrac{1}{\sqrt2}\right)$.
Work Step by Step
Differentiate $y^4 + 1 = y^2 + x^2$ with respect to $y$ using chain rule.
We get, $4y^3=2y+2x\dfrac{dx}{dy}$
Now solve for $\dfrac{dx}{dy}$.
$2x\dfrac{dx}{dy}+2y=4y^3$
$\implies 2x\dfrac{dx}{dy}=4y^3-2y$
$\implies \dfrac{dx}{dy}=\dfrac{4y^3-2y}{2x}$
$\implies \dfrac{dx}{dy}=\dfrac{2y^3-y}{x}$
Since, tangent line is vertical when $\dfrac{dx}{dy}=0$.
Substitute $\dfrac{dx}{dy}=0$ in $\dfrac{dx}{dy}=\dfrac{2y^3-y}{x}$.
We get, $0=\dfrac{2y^3-y}{x}$
Since, $x\ne0$
$\implies 2y^3-y=0$
$\implies y(2y^2-1)=0$
$\implies y=0$ or $2y^2-1=0$
Now solve $2y^2-1=0$ as follows.
$2y^2-1=0$
$\implies y^2=\dfrac{1}{2}$
$\implies y=\pm\dfrac{1}{\sqrt2}$
So, at $y=0$, $y=\dfrac{1}{\sqrt2}$ and $y=-\dfrac{1}{\sqrt2}$ the tangent line is vertical.
To find the x-coordinates substitute values of y-coordinate in $y^4 + 1 = y^2 + x^2$.
For $y=0$ x-coordinates can be calculated as follows.
$0^4 + 1 = 0^2 + x^2$
$\implies x^2=1$ or $x=\pm1$
So the points with $y=0$ are $(1,0)$ and $(-1,0)$.
For $y=\dfrac{1}{\sqrt2}$ x-coordinates can be calculated as follows.
$\left(\dfrac{1}{\sqrt2}\right)^4 + 1 = \left(\dfrac{1}{\sqrt2}\right)^2 + x^2$
$\implies \dfrac{1}{4} + 1 = \dfrac{1}{2} + x^2$
$\implies x^2=\dfrac{1}{4}+1-\dfrac{1}{2}=\dfrac{1+4-2}{4}=\dfrac{3}{4}$
$\implies x=\pm\dfrac{\sqrt{3}}{2}$
So the points with $y=\dfrac{1}{\sqrt2}$ are $\left(\dfrac{\sqrt{3}}{2},\dfrac{1}{\sqrt2}\right)$ and $\left(-\dfrac{\sqrt{3}}{2},\dfrac{1}{\sqrt2}\right)$.
For $y=-\dfrac{1}{\sqrt2}$ x-coordinates can be calculated as follows.
$\left(-\dfrac{1}{\sqrt2}\right)^4 + 1 = \left(-\dfrac{1}{\sqrt2}\right)^2 + x^2$
$\implies \dfrac{1}{4} + 1 = \dfrac{1}{2} + x^2$
$\implies x^2=\dfrac{1}{4}+1-\dfrac{1}{2}=\dfrac{1+4-2}{4}=\dfrac{3}{4}$
$\implies x=\pm\dfrac{\sqrt{3}}{2}$
So the points with $y=\dfrac{1}{\sqrt2}$ are $\left(\dfrac{\sqrt{3}}{2},-\dfrac{1}{\sqrt2}\right)$ and $\left(-\dfrac{\sqrt{3}}{2},-\dfrac{1}{\sqrt2}\right)$.
Thus, the points where the tangent line is vertical are $(1,0)$, $(-1,0)$, $\left(\dfrac{\sqrt{3}}{2},\dfrac{1}{\sqrt2}\right)$, $\left(-\dfrac{\sqrt{3}}{2},\dfrac{1}{\sqrt2}\right)$, $\left(\dfrac{\sqrt{3}}{2},-\dfrac{1}{\sqrt2}\right)$ and $\left(-\dfrac{\sqrt{3}}{2},-\dfrac{1}{\sqrt2}\right)$.
