The line tangent to a curve, no derivative

[icantadd]

Homework Statement

Let (x₀,y₀) be a point on the graph f(x) = x² for x₀ not equal to 0. Find the equation of the line tangent to the graph of f at that point by finding a line that intersects the cure in exactly one point. Do not use the derivative to find this line.

Homework Equations

f(x) = x²
y=mx+b
(y₁-y₂) = m(x₁-x₂)

The Attempt at a Solution

The way I would normally approach this problem is to use the derivative. But that is a syntax error in this problem; however, it does not stop me from knowing that the slope should depend on x₀, where the slope of some x₀ should be 2*x₀.

At the same time, any point on the graph can be given in the form (x₀,(x₀)²). I tried to force through on this point, and get an equation (y- (x₀)²) = m(x-x₀). I would think I need to solve for m, but then there are too many variables, and I get too many possible solutions.

Also, I tried to set x² = m*x + b, to see the set of solutions. That is unhelpful, there are solutions to that set of equations that are not tangents to the graph, but instead are secants.

I believe that my general approach to the problem is wrong; I am missing something. I would love any advice.

 

[Dick]

Your second approach is the right one. The curve is y=x^2, the line is y=m*(x-x0)+x0^2 (by forcing it to go through the right point as you did in the first part). So set x^2=m*(x-x0)+x0^2. Solve for x. For a general value of m, you will get two solutions (one, of course, is x=x0). What's the other one? Now find the value of m where those two solutions are equal.

 

[icantadd]

Thank you sir.

Playing with the algebra starting with x^2 = m*(x-x0) +x0^2, I get:

(x-x0)(x+x0)=m*(x-x0), which simplifies to

m = x + x0 <=> x = m -x0. The other solution of x is x = x0 given by Dick.

Setting the two equal to each other is:

m - x0 = x0 Therefore,

m = 2*x0.

Which is what the answer should be. Brilliant, thank you again.

Question: To get that one solution is x = x0, did you notice that both the left and right hand sides of the equation in
(x-x0)(x+x0) = m*(x-x0) are equal if both are 0, which happens when x=x0, or is there another way to get that?

 

[Dick]

Thank you sir.

Playing with the algebra starting with x^2 = m*(x-x0) +x0^2, I get:

(x-x0)(x+x0)=m*(x-x0), which simplifies to

m = x + x0 <=> x = m -x0. The other solution of x is x = x0 given by Dick.

Setting the two equal to each other is:

m - x0 = x0 Therefore,

m = 2*x0.

Which is what the answer should be. Brilliant, thank you again.

Question: To get that one solution is x = x0, did you notice that both the left and right hand sides of the equation in
(x-x0)(x+x0) = m*(x-x0) are equal if both are 0, which happens when x=x0, or is there another way to get that?

That, and that you deliberately set the tangent equation up so that at x=x0, the two would be equal.

 

[HallsofIvy]

This was, by the way, Fermat's method for finding tangents.