Proof of x and y int of a line

[lax1113]

Homework Statement

prove that $$\sqrt{x}$$ + $$\sqrt{y}$$ = c
Show that the sum of the x and y intercepts of any tangents to the line above = c (some positive constant).

Homework Equations

y1 - y2 = m(x1 - x2)
dy/dx(for this problem) = -$$\sqrt{y}$$/$$\sqrt{x}$$

The Attempt at a Solution

So I get the slope, as written above, and put it into a point/slope equation, but where from here? When I try to solve for y = 0 and x = 0 I always have y2 and x2 left, I think I might just be doing something completely wrong, I haven't done something like this for a while. Is this even the right direction? Solve for the y and x int by making the opposite 0 in the equation, and then try to get the results, (the two interecepts) to add up to be $$\sqrt{x}$$ + $$\sqrt{y}$$

 

[HallsofIvy]

First problem, I don't know what "x1", "y1", "x2", and "y2" are since you don't say. You mean, I think, that the equation of the tangent line at $(x_1, y_1)$ is [itex]y- y_1= m(x- x_1).

But your real problem is that the derivative of y with respect to x is NOT $-\sqrt{y}/\sqrt{x}$. You have x and y reversed.

 

[Architeuthis Dux]

= c

In order to prove that the sum of the x and y intercepts of any tangents to the line \sqrt{x} + \sqrt{y} = c is equal to c, we need to use the definition of tangents and the properties of lines.

First, let's define a tangent line as a line that intersects a curve at exactly one point and has the same slope as the curve at that point. In this case, the curve is the line \sqrt{x} + \sqrt{y} = c.

Next, we can use the point-slope form of a line to find the equation of a tangent line to this curve. Since we know that the slope of the curve at any point is given by dy/dx = -\sqrt{y}/\sqrt{x}, we can use the point-slope form to find the equation of the tangent line at any given point (x,y).

So, the equation of the tangent line at (x,y) is y - y = -\sqrt{y}/\sqrt{x}(x - x). Simplifying this, we get y = -\sqrt{y}/\sqrt{x}x + y.

Now, we can use the definition of intercepts to find the x and y intercepts of this tangent line. The x intercept occurs when y = 0, so we can substitute this into our equation to get x = -\sqrt{y}/\sqrt{x}x. Solving for x, we get x = 0.

Similarly, the y intercept occurs when x = 0, so we can substitute this into our equation to get y = -\sqrt{y}/\sqrt{x}(0) + y. Simplifying, we get y = y.

So, the tangent line intersects the x axis at x = 0 and the y axis at y = y. The sum of these intercepts is 0 + y = y.

Now, we can use the original equation \sqrt{x} + \sqrt{y} = c to substitute for y in terms of x. Solving for y, we get y = (c - \sqrt{x})^2.

Substituting this into our equation for the y intercept, we get y = -\sqrt{(c - \sqrt{x})^2}/\sqrt{x}x + (c - \sqrt{x})^2.

Simplifying, we get y = -c