# Thread: Find the slope of a line tangent to a curve.

1. ## Find the slope of a line tangent to a curve.

I encountered this problem months ago in a math competition, and have yet to have any understanding how it could be solved.

A non-vertical line with y-intercept 2 is tangent to the curve given by the equation x^2 + y^2 = 12x + 20y - 100. Find the slope of the line.

This problem would be easy if it was single variable, but because it has both x and y, I have zero clue where to start.

2. ## Re: Find the slope of a line tangent to a curve.

$x^2+y^2=12x+20y-100$. Let's use the chain rule: $\frac{d}{dx}\left[x^2+y^2\right]=2x+2y\frac{dy}{dx}=\frac{d}{dx}\left[12x+20y-100\right]=12+20\frac{dy}{dx}$. Solving for $\frac{dy}{dx}$, we get $\frac{dy}{dx}=\frac{12-2x}{2y-20}=\frac{6-x}{y-10}$. Can you take it from there?

Edit: My apologies. I didn't see that this was in Pre-Calculus. If you haven't learned derivatives yet, you'll need some trigonometry.

3. ## Re: Find the slope of a line tangent to a curve.

Start by figuring out the circle's characteristics. Then you can finish with a bit of trigonometry.

4. ## Re: Find the slope of a line tangent to a curve.

In other words, it is an "implicit function"- y is defined "implicitly" by that formula- you could, theoretically, solve for y.

There are two ways to continue. One, the simpler in concept but harder in practice, is to actually solve for y as a function of x: $x^2+ y^2= 12x + 20y -100$ can be written as $y^2- 20y+ x^2- 12x+ 100= 0$ and now treat that as a quadratic equation in y with x just some number: by the quadratic formula, $y= \frac{-20\pm\sqrt{400- 4(x^2- 12x+ 100}}{2}= \frac{-20\pm\sqrt{48x- 4x^2}}{2}= -10\pm\sqrt{12- x^2}$ and differentiate that.

The other way, more "sophisticated" but easier in practice, is to use "implicit differentiation". By the chain rule the derivative of $y^2$ with respect to x is the derivative of $y^2$ with respect to y times the derivative of y with respect to x: 2yy'. And the derivative of 20y with respect to x is 20y'. Differentiating both sides of $x^2+ y^2= 12x+ 20y- 100$ with respect to x gives $2x+ 2yy'= 12+ 20y'$. Solving for y', $20y'- 2yy'= (20-2y)y'= 2x- 12$, $y'= \frac{2x- 12}{20- 2y}= \frac{x- 6}{10- y}$.

In either case, you still have the problem that you do not know at what "(x,y)" you want to have the tangent line. Instead, replace x with some letter, say, a, and y with its value in terms of that "a". Write the equation of the tangent plane and set the value of y, when x= 0, equal to 2. The solve that equation for a.

5. ## Re: Find the slope of a line tangent to a curve.

I would let the tangent line be:

$y=mx+2$

With the given curve, complete the square on the two variables to obtain:

$(x-6)^2+(y-10)^2=6^2$

Now, substitute for $y$ into the equation, then require the discriminant to be zero (do you understand why?). Now solve for $m$.

6. ## Re: Find the slope of a line tangent to a curve.

I do know derivatives, but I put it in precalculus because I knew they would not be complex for this problem. I got (6-x) / (y-10) as the slope, however, that is not enough information to find the slope as we have no point to give, and therefore cannot create a slope when the only info on the tangent line I have is that it takes the form y = mx + 2. I'm still exploring the thought of trigonometry.

7. ## Re: Find the slope of a line tangent to a curve.

Your BIG hint for today is to consider what $m$ is. You'll have two equations and two unknown variables.