SOLUTION: Find all points on the circle x^2 + y^2 = 81 where the slope is 9/40.
(x, y) = ( , ) (positive y-coordinate)
(x, y) = ( , ) (negative y-coordinate)
I've been trying everyt
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-> SOLUTION: Find all points on the circle x^2 + y^2 = 81 where the slope is 9/40.
(x, y) = ( , ) (positive y-coordinate)
(x, y) = ( , ) (negative y-coordinate)
I've been trying everyt
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Question 1157854: Find all points on the circle x^2 + y^2 = 81 where the slope is 9/40.
(x, y) = ( , ) (positive y-coordinate)
(x, y) = ( , ) (negative y-coordinate)
I've been trying everything and attempting to solve based on similar questions, please include all steps so I also know how to solve afterwards Answer by Edwin McCravy(20054) (Show Source):
That means:
Find all points on the circle x^2 + y^2 = 81 where the slope OF A LINE DRAWN
TANGENT TO THE CIRCLE AT THAT POINT is 9/40.
We take the derivative implicitly
Divide every term by 2
Substitute 9/40 for (dy)/(dx)
Solve for y:
Multiply both sides by 40/9
Substitute for y in the equation of the circle:
Multiply through by 81
Substituting for x² in the equation of the circle:
Multiply through by 1681
So the two points where the slope of the tangent line is 9/40 are:
and
Edwin
