Question 1124567: Find the equation of tangent line at the point (2,1) on the circle 2x^2+2y^2=5x.
Found 2 solutions by htmentor, Alan3354:
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! From implicit differentiation, we get:
4x + 4y(dy/dx) = 5
Thus the slope of the tangent line, dy/dx = (5 - 4x)/4y
Using the point slope form and x = 2, y = 1, we get:
y - 1 = ((5-8)/4)(x - 2)
Simplify:
y = (-3/4)x + 10/4
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! Find the equation of tangent line at the point (2,1) on the circle 2x^2+2y^2=5x.
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Find the slope at (2,1)
2x^2+2y^2=5x
4x*dx + 4y*dy = 5dx
(4x-5)*dx = -4y*dy
dy/dx = -(4x-5)/4y
@(2,1) dy/dx = -3/4
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y-1 = -(3/4)*(x-2)
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If you want to do it without calculus email via the TY note, or post it again and ask for an algebraic solution.
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