. Let's use the chain rule: . Solving for , we get . 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.
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.
. Let's use the chain rule: . Solving for , we get . 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.
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: can be written as and now treat that as a quadratic equation in y with x just some number: by the quadratic formula, and differentiate that.
The other way, more "sophisticated" but easier in practice, is to use "implicit differentiation". By the chain rule the derivative of with respect to x is the derivative of 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 with respect to x gives . Solving for 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.
I would let the tangent line be:
With the given curve, complete the square on the two variables to obtain:
Now, substitute for into the equation, then require the discriminant to be zero (do you understand why?). Now solve for .
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.