What is the slope of the tangent line of # x/(x-3y) = C #, where C is an arbitrary constant, at #(-2,1)#?

Question

Evelyn Agard · Accepted Answer

The given equation represents the straight line

#y = (1-1/C)/3x#. As C is fixed as 2/5, by setting (-2, 1) on the line,

the equation is #y = -x/2#. Its slope is #-1/2#.

Talking about tangent to a straight line.is tautology.

Emma Aldridge · Answer

undefined To find the slope of the tangent line at the point (-2, 1), we first need to find the derivative of the function with respect to both x and y. Then, we can evaluate it at the given point (-2, 1) to find the slope of the tangent line.

Differentiating the given equation $ \frac{x}{x - 3y} = C $ with respect to x gives:

$$ \frac{\partial}{\partial x} \left( \frac{x}{x - 3y} \right) = \frac{\partial}{\partial x} (C) $$

$$ \frac{x - (x - 3y) \cdot 1}{(x - 3y)^2} = 0 $$

$$ \frac{3y}{(x - 3y)^2} = 0 $$

Differentiating with respect to y gives:

$$ \frac{\partial}{\partial y} \left( \frac{x}{x - 3y} \right) = \frac{\partial}{\partial y} (C) $$

$$ \frac{-3x}{(x - 3y)^2} = 0 $$

Now, to find the slope of the tangent line at (-2, 1), substitute x = -2 and y = 1 into the derivative expressions for $\frac{\partial x}{\partial y}$ and $\frac{\partial y}{\partial x}$.

$$ \frac{-3(1)}{((-2) - 3(1))^2} = \frac{-3}{(-2 - 3)^2} = \frac{-3}{(-5)^2} = \frac{-3}{25} $$

Thus, the slope of the tangent line at the point (-2, 1) is $ \frac{-3}{25} $.