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

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?
 
Thread 'Direction Fields and Isoclines'

Thread 'Direction Fields and Isoclines'

I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
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