# Finding points on graph with tangent lines perpendicular to a line

Find all points $(x,y)$ on the graph of $y=\frac{x}{x-3}$ with tangent lines perpendicular to the line $y=3x-1.$

My thoughts on this problem:

First I should find the slope of the given line and the tangent to the given curve. I'm unsure of how to proceed with this though. I know that the slope of the tangent line is equal to $\frac{dy}{dx}$ at any point on the curve.

So the slope of the tangent line would be: $$y'=\frac{(x-3)(1)-(x)(1)}{(x-3)^2}=\frac{-3}{(x-3)^2}$$

I also know that the product of the slopes of two perpendicular lines is $-1.$ I'm not sure how to apply this to the problem, and also I'm not sure about how to find $x$- and $y$-coordinates.

Overall I'm not sure about how to set this problem up step by step. Thank you.

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The slope of a line perpendicular to $y = mx + b$ is equal to $-\frac 1m$, where slope $m$ of the equation $y = 3x - 1$ is $m = 3$.

So solve for the points at which the first derivative is equal to $-\dfrac 1m = -\dfrac 13$. You'll get a quadratic equation, from which you can find the two solutions.

$$y'=\frac{x-4}{(x-3)^2} = -\dfrac 13$$

$$(x - 3)^2 = -3( x- 4),\;\;x \neq 3$$ $$x^2 - 6x + 9 = -3x + 12$$ $$\iff x^2 - 3x - 3 = 0$$

Those solutions, (solving for $x$), will give you the $x$-coordinate of the points at which the tangent lines to the curve are perpendicular to the given line. We can simply use the quadratic equation to find the solutions, $x_1$ and $x_2$.

To find the corresponding $y$ coordinate, just plug each solution $x_i$ into the original curve (not into $y'$) and compute $y$.

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Thank you! If possible, could you show some more steps? –  user437158 Nov 19 '13 at 0:58
@amWhy: I finished mine, but still not awake! :-) –  Amzoti Nov 19 '13 at 14:09
@amWhy thank you for added steps! I was multiplying incorrectly. –  user437158 Nov 19 '13 at 14:42
You're welcome, @user436158! –  amWhy Nov 19 '13 at 14:47