Find the equation of the tangent lines to the following curve at the indicated point.
y= x / y+a at (0,0)
Thanks.
Is this
$\displaystyle y = \frac{x}{y + a}$ at (0, 0)?
The solution method is to solve for the derivative at the point x = 0. This then is the slope of your tangent line, and you may use some version of the line equation and the given point to find the equation for the line.
In this case we would probably want to use implicit differentiation:
$\displaystyle \frac{dy}{dx} = \frac{1(y + a) - x \cdot \frac{dy}{dx} }{(y + a)^2}$
So
$\displaystyle (y + a)^2 \frac{dy}{dx} = y + a - x \cdot \frac{dy}{dx}$
$\displaystyle [(y + a)^2 + x] \frac{dy}{dx} = y + a$
$\displaystyle \frac{dy}{dx} = \frac{y + a}{(y + a)^2 + x}$
At the point (0, 0) the value of the derivative is
$\displaystyle \frac{dy}{dx} = \frac{0 + a}{(0 + a)^2 + 0} = \frac{1}{a}$
So the slope of the tangent line at (0, 0) is 1/a. Thus the tangent line has the form:
$\displaystyle y = \frac{1}{a} \cdot x + b$
We know this line passes through the point (0, 0), so
$\displaystyle 0 = \frac{1}{a} \cdot 0 + b$
Thus b = 0 and the tangent line is
$\displaystyle y = \frac{1}{a} \cdot x$
-Dan
Hello, Paige05!
Find the equation of the tangent lines (?) to the following curve at the indicated point.
. . $\displaystyle y \:= \:\frac{x}{y+a}$ at $\displaystyle (0,0)$
We have: .$\displaystyle y^2 + ay \:=\:x$
Differentiate implicitly: .$\displaystyle 2yy' + ay' \:=\:1\quad\Rightarrow\quad y'(2y+a) \:=\:1\quad\Rightarrow\quad y' \:=\:\frac{1}{2y + a}$
At $\displaystyle (0,0)\!:\;\;y' \:=\:\frac{1}{a}$
The equation of the tangent is: .$\displaystyle y - 0 \:=\:\frac{1}{a}(x-0)\quad\Rightarrow\quad\boxed{ y \:=\:\frac{1}{a}x}$