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Consider the curve C given by the equation

Consider the curve C given by the equation: C · · · √ x + √ y = √ a, where a is a constant, a > 0. Let (x0, y0) with x0 > 0, y0 > 0, be a point on C. If (x1, 0) and (0, y1) are x and y intercepts of the tangent line to C at (x0, y0), show that x1 + y1 = a.
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Please state your question correctly.  The reason why you did not get an answer yesterday is because your equation is unclear to decipher.
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A good strategy for solving this problem might be to make up a specific example and think about how you would go about it then (for example let a=36). That said, we can write the equation of the tangent line passing through (x0, y0) by finding y' (implicitly), getting the slope, and then using point-slope. After implicit differentiation y' = -(y/x)1/2, so the slope at the given point is -(y0/x0)1/2. Call the value of the slope m so I do not have to keep typing it. That means the equation of the tangent line is y = m(x - x0) + y0. When x=0, y = y1. Make those substitutions and solve for y1. When y = 0, x = x1. Make those substitutions and solve for x1. Add the expressions for x1 and y1. The resulting expression will be in terms of x0 and y0. Now here is the key. Since (x0, y0) lies on the curve it must satisfy the original equation. That is (x0)1/2 + (y0)1/2 = a1/2. Try squaring both sides of that equation to get "a" in terms of x0 and y0. You will see that that expression is equal to the expression you got for x1 + y1. Quite a bit of algebra involved in all of this. I suggest using fractional exponents instead of radicals as you work through it. Take a look at this specific exampe: https://www.desmos.com/calculator/4gnvaxlu5b Try the slider to see the tangent moving. But x1 + y1 always is the same value.