calculusderivativesimplicit-differentiation
first time poster so sorry if I'm doing something wrong.
"Consider the closed curve in the xy-plane given by $2x^2 – xy + y^3 + x = 9$.
Find equation(s) of all tangent lines to the curve at $y = 1$."
So after I took the derivative I got:
$$\frac{y – 4x – 1}{3y^2 – x}$$
After that I have no clue what I'm supposed to do next. Sorry about the formatting and thanks for the help.
The derivative is correct.
As stated in the execise, the point $(1,2)$ is not on the curve; so you need to find such a point $(x_0,y_0)$ which a) lies on the curve; b) the tangent line to the curve in this point passes through $(1,2)$. Are these hints sufficient?
Edit
Our line is described by the equation $y=kx+c$. It is tangent to the curve int he point $(x_0,y_0)$, thus $y_0=kx_0+c$; in addition, the derivatives of the line and of the curve must coincide in this point, i.e. $k=\frac{1}{(x_0+1)^2}$. Finally, the line passes though $(1,2)$, hence $2= k+c$.
Thus, we obtain the equations: $$y_0=kx_0+c\\y_0=x_0/(x_0+1)\\k=\frac{1}{(x_0+1)^2}\\2= k+c$$
Can you solve these equations to obtain all possible pairs $(k,c)$?
Pick some point $(a,b)$ which lies on the curve $y=\frac{x}{x+1}$. This means that $b=\frac{a}{a+1}$. Taking the derivative using the quotient rule, we find:
$$y'(x)=\frac{1}{(1+x)^2}$$
so the slope of the tangent line at our point $(a,b)$ is
$$y'(a)=\frac{1}{(1+a)^2}$$
The point slope formula then tells us that the equation of the tangent line to the curve at the point $(a,b)$ is
$$y-b=\frac{1}{(1+a)^2}(x-a)$$
We want the point $(1,2)$ to satisfy this equation, so remembering that $b=\frac{a}{a+1}$, we have:
$$2-b=2-\frac{a}{a+1}=\frac{1}{(1+a)^2}(1-a)$$
This equation reduces to the quadratic $a^2+4a+1=0$ (which is the quadratic you found). The solutions $a=-2\pm\sqrt{3}$ can be plugged back into the equation
$$y=\frac{1}{(1+a)^2}(x-a)+\frac{a}{1+a}=\frac{1}{(1+a)^2}x+\frac{a^2}{(1+a)^2}$$
to obtain the equations of the desired tangent lines.
Here is a helpful picture plotted with Wolfram|Alpha:
Notice where the tangent lines intersect!
Best Answer
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:
$$y - 1 = m_i(x - x_i)$$