Proof of x and y int of a line

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SUMMARY

The discussion centers on proving that the sum of the x and y intercepts of tangents to the line defined by the equation \(\sqrt{x} + \sqrt{y} = c\) equals a constant \(c\). Participants clarify that the slope of the tangent line is derived from the derivative \(\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}\), but there is confusion regarding the correct application of the point-slope form of the equation. A critical error identified is the reversal of variables in the derivative, which must be corrected for accurate calculations.

PREREQUISITES
  • Understanding of calculus, specifically derivatives and slopes of tangent lines.
  • Familiarity with the point-slope form of a linear equation.
  • Knowledge of intercepts in coordinate geometry.
  • Ability to manipulate square roots and algebraic expressions.
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  • Review the concept of derivatives and their geometric interpretations in calculus.
  • Study the point-slope form of linear equations and its applications in finding tangent lines.
  • Explore the properties of intercepts in relation to curves and lines.
  • Practice solving problems involving square roots and their implications in algebraic equations.
USEFUL FOR

Students studying calculus, particularly those tackling problems involving derivatives, tangent lines, and intercepts. This discussion is beneficial for anyone looking to strengthen their understanding of these concepts in mathematical analysis.

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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}
 
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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 y- y_1= m(x- x_1). <br /> <br /> 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.
 

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