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

[delriofi]

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?

 

[micromass]

Hi delriofi!

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).

 

[delriofi]

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?

 

[Hurkyl]

This is a duplicate of this post:

https://www.physicsforums.com/showthread.php?p=3330285#post3330285


I'm closing this one.

 

[blue_raver22]

I would first clarify that the equation given is for an ellipse with center at the origin and major and minor axes of length 6 and 8, respectively. This information is important for understanding the geometry of the ellipse and finding its points of tangency.

To find the points on an ellipse where the slope of the tangent line equals 1, we can use the fact that the slope of the tangent line at a point on the ellipse is given by the derivative of the ellipse equation with respect to y, divided by the derivative of the ellipse equation with respect to x. In other words, the slope of the tangent line at a point (x,y) on the ellipse is given by dy/dx = -16x/9y.

To find the points where the slope of the tangent is 1, we can set dy/dx = 1 and solve for the x and y values that satisfy this equation. In this case, we get the equation -16x/9y = 1, which simplifies to -16x = 9y. This is a linear equation in x and y, and we can solve for one variable in terms of the other. For example, if we solve for x, we get x = (-9/16)y.

Substituting this value of x into the equation for the ellipse, we get (-9/16)y^2/9 + y^2/16 = 1, which simplifies to 7y^2/144 = 1. Solving for y, we get y = ±12/√7. Substituting this value of y into the equation for x, we get x = ±(9/16)(12/√7) = ±27/4√7.

Therefore, the points on the ellipse where the slope of the tangent line equals 1 are (27/4√7, 12/√7) and (-27/4√7, -12/√7). These points can be verified by substituting them into the derivative dy/dx = -16x/9y and confirming that it is equal to 1.

In conclusion, to find the points on an ellipse where the slope of the tangent line equals 1, we can set the derivative of the ellipse equation equal to 1 and solve for the x and y values that satisfy this equation. This approach can be applied to any