Plug in $y = 1$ to your original equation: $2x^2 - xy + y^3 + x = 9$ and obtain a quadratic equation in $x$, which can be solved. At $y = 1$, we have $$2x^2 - x(1) + (1)^3 + x = 9\iff 2x^2 - 8 = 0 \iff x^2 - 4 = (x - 2)(x + 2) = 0$$ So we have two points on the curve where $y = 1$: one point corresponding to each solution: $x_1 = 2$, $x_2 = -2$. That is, you need to find the equations of the tangent line at the points $(2, 1)$ and the tangent line to the curve at $(-2,1)$.

Use $\dfrac{dy}{dx}= \dfrac{y - 4x - 1}{3y^2 - x}$, which you've already found, to compute the slope of tangent line at each point; i.e., evaluate $\frac {dy}{dx}$ at the point $(2, 1)$ to find $m_1$, and again at the point $(-2, 1)$ to find $m_2$.

$$\text{At}\;\;(2, 1),\quad\; m_1= -8$$

$$\text{At}\;\;(-2, 1),\quad m_2 = \dfrac {8}{5}$$

Then you'll have, for each point you've found $(x_i, 1)$, and the corresponding slope $m_i$ of the tangent line, a point on the line, and its slope. So you can write the corresponding equation of the tangent line using point-slope form of the equation of a line:

$$y - 1 = m_i(x - x_i)$$

The slope of the tangent line at $(1,2)$ being $0$ implies it is the horizontal line $y=2$.

The normal line is perpendicular to the tangent line at $(1,2)$, hence it is the vertical line $x=1$.

Refer to the graph:

$\hspace{1cm}$

## Best Answer

Your work is correct, and it seems that your suggested answer is wrong. When the value of the slope is undefined, this points to the possibility that the tangent line is vertical. This is the case for your problem, so the tangent line should be x=2.

To see this more rigorously, you could consider $x$ as a function of $y$ and calculate $\frac{dx}{dy}$ at $(2,-3)$. You should get $0$. Then use this 'gradient' in the equation $m(y-y_1)=x-x_1$.

If the answer you've listed is correct, perhaps you have copied the problem down incorrectly.