Finding the points on an ellipse where the slope of the tangent line equals 1

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SUMMARY

The discussion focuses on finding points on the ellipse defined by the equation x²/9 + y²/16 = 1 where the slope of the tangent line equals 1. The slope is given by the derivative dy/dx = -16x/9y. By setting this equal to 1, the equation 1 = -16x/9y is derived, leading to the substitution 9y = -16x. However, attempts to substitute this back into the ellipse equation resulted in a contradiction, indicating a potential misunderstanding or miscalculation in the approach.

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delriofi
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If there is an an ellipse x^2/9 + y^2/16 = 1, and the slope of the tangent is dx/dy = -16x/9y, how do you find what points at which the slope of the tangent is 1? I have no idea how to answer this and I've been trying for like an hour. Can anyone help me?
 
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Hi delriofi! :smile:

delriofi said:
If there is an an ellipse x^2/9 + y^2/16 = 1, and the slope of the tangent is dx/dy = -16x/9y, how do you find what points at which the slope of the tangent is 1? I have no idea how to answer this and I've been trying for like an hour. Can anyone help me?

Don't you mean dy/dx=-16x/9y ?
Well, since you know the slope is 1, you know that 1=-16x/9y, thus 9y=-16x. Substitute that back into the equation for the ellipse and solve for x (or y).
 
Ok so I did x^2/9 + (-9y/16)^2/16 = 1 and I get that x^2/9 - x^2/9 = 1 but that doesn't make sense does it?
 

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