Calculus/Slope of the Tangent
Expert: Paul Klarreich - 12/8/2009
Questionprove that the sum of two intercept on the axes of any tangent to the curve x^1/2+y^1/2=a^1/2 is constant
AnswerQuestioner: anu
Country: Pakistan
Category: Calculus
Private: No
Subject: plz help
Question: prove that the sum of two intercept on the axes of any tangent to the curve
x^1/2+y^1/2=a^1/2 is constant
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Step 1: Find dy/dx.
Step 2: Let the point of tangency be at x = x0, and find m = y'(x0)
Step 3: write the equation of the line using:
y - y0 = m(x - x0)
Step 4:
Find the y-intercept by setting x = 0.
Find the x-intercept by setting y = 0.
Step 4: Write their sum and finish the proof.
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Abbreviations: A = sqrt(a), sx = x^1/2 = sqrt(x), s(x0) = sqrt(x0)
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If x^1/2 + y^1/2 = a^1/2, then
y = (A - sx)^2
Step 1: ---------------
y' = 2(A - sx)(-1/2sx)
y' = (A - sx)(-1/sx), which is indeed negative, because sqrt(x) < sqrt(a)
y' = (sx - A)/sx, which is indeed negative, because sqrt(x) < sqrt(a)
y' = 1 - A/sqrt(x)
Step 2: ---------------
Now let the point of tangency be at x = x0.
m = 1 - A/sqrt(x0)
Step 3:---------------
y0 = (A - s(x0))^2
Equation is
y - y0 = m(x - x0)
y - (A - s(x0))^2 = (1 - A/s(x0))(x - x0)
Step 4:---------------
Get the y-int:
y - (A - s(x0))^2 = (1 - A/s(x0))(- x0)
y = (A - s(x0))^2 + (1 - A/s(x0))(- x0)
y = (A - s(x0))^2 - x0 + As(x0)
y = A^2 + x0 - 2A s(x0) - x0 + As(x0)
y = A^2 - A s(x0) <<<<<<<<<<<<<<<<<<<<<< y-intercept
Get the x-int: (set y = 0)
0 - (A - s(x0))^2 = (1 - A/s(x0))(x - x0)
- (A^2 + x0 - 2As(x0)) = (1 - A/s(x0))(x - x0)
- A^2 - x0 + 2As(x0) = (1 - A/s(x0))(x - x0)
Mult through by sx0:
- A^2 sx0 - x0 s(x0) + 2A x0 = (s(x0) - A)(x - x0)
- A^2 sx0 - x0 s(x0) + 2A x0 = s(x0)x - Ax - x0 s(x0) + Ax0
- A^2 sx0 + A x0 = s(x0)x - Ax
Factor:
A s(x0)(- A + s(x0)) = (s(x0) - A)x
Cancel:
A s(x0) = x <<<<<<<<<<<<<<<<<< x-intercept
Step 5: ---------------
FINALLY, add them:
x+y = A^2 - A s(x0) + A s(x0) = A^2, WHICH IS INDEPENDENT OF x0