# I on Finding the tangent of the ellipse

• Dave J
In summary, the conversation is about finding the equation of a line with a positive slope that is tangent to the given ellipse at x=2. The steps involve finding the derivative of the ellipse equation, solving for dy/dx, and using the point (x=2, y=0.37) to find the slope of the tangent line. However, there is confusion about the correct value for y, with one solution being y = sqrt(20)/3 and the other being y = -sqrt(20)/3. The correct solution is determined to be y = -sqrt(20)/3, which results in a positive slope for the tangent line.
Dave J

## Homework Statement

The problem is

Find the equation of the line with a positive slope that is tangent to the ellipse
(x^2)/9 + (y^2)/4 = 1
At x=2

## Homework Equations

Now I know that to find the tangent, I find the derivative of the equation. So I got
2x/9 + 2y/4 dy/dx = 0But its this part where I can't go any further. See I only get X, so I can't solve for dy/dx without having y in the answer, and therefor I can't solve for y.

I solved for dy/dx and got -16/18y, and just tp test I found what was Y when x = 2 for the original equation, it was 0.37.
So I adding 0.37 into -16/18y, i got -2.4 (which would be the slope of the line)

But it says find the line with the positive slope, after reading through the entire chapter in my textbook, and searching the internet, I am stuck and don't know what I'm doing wrong.

Last edited:
it should be be 2x/9 + 2y/4 dy/dx = 0

If x = 2, then (y^2)/4 = 1 - 4/9

solve for y.

Last edited:
it should be be 2x/9 + 2y/4 dy/dx = 0

Sorry that was a typo, I originally had that, and typed it out wrong.

If x = 2, then (y^2)/4 = 1 - 4/9

solve for y.

If x = 2, then (y^2)/4 = 1 - 4/9

solve for y.

I did solve for y, it was 0.37 but my answer was negative and therefor couldn't be it.

y = sqrt(20)/3 or -sqrt(20)/3dy/dx = -8x / 18y

and y = -sqrt(20)/3so it will be positive

Last edited:
and y = -sqrt(20)/3

Where did you get that from. When I solved for y in the original equation. I get

0.4444 + (y^2)/4 = 1
0.55556 / 4 = y^2

Sqrt(0.138) = y

y = 0.37

$$\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$$

$$\frac{y^{2}}{4} = 1 - \frac{4}{9} = \frac{5}{9}$$

$$y^{2} = \frac{20}{9}$$

$$y = \pm \frac{\sqrt{20}}{3}$$

Dave J said:
and y = -sqrt(20)/3

Where did you get that from. When I solved for y in the original equation. I get

0.4444 + (y^2)/4 = 1
0.55556 / 4 = y^2
No, y^2= 4(0.55556)= 2.22222
so y= sqrt(2.22222)= 1.4907= sqrt(20)/3

Sqrt(0.138) = y

y = 0.37

## 1. What is the equation for finding the tangent of an ellipse?

The equation for finding the tangent of an ellipse is y = mx + b, where m is the slope of the tangent line and b is the y-intercept.

## 2. How do you determine the slope of the tangent line on an ellipse?

The slope of the tangent line on an ellipse can be determined using the formula m = -b / a, where a and b are the major and minor axes of the ellipse, respectively.

## 3. Can the tangent line intersect the ellipse at more than one point?

No, the tangent line can only intersect the ellipse at one point. This is because the tangent line is drawn perpendicular to the radius at that point, and the radius can only intersect the ellipse at one point.

## 4. How do you use derivatives to find the tangent of an ellipse?

To find the tangent of an ellipse using derivatives, you would first find the derivative of the ellipse equation. Then, substitute the x-coordinate of the point of tangency into the derivative to find the slope of the tangent line.

## 5. Are there any special cases for finding the tangent of an ellipse?

Yes, there are two special cases for finding the tangent of an ellipse: when the ellipse is a circle (in which case the tangent line is perpendicular to the radius at all points), and when the ellipse is a parabola (in which case the tangent line is parallel to the directrix at all points).

• Calculus and Beyond Homework Help
Replies
4
Views
244
• Calculus and Beyond Homework Help
Replies
4
Views
861
• Calculus and Beyond Homework Help
Replies
1
Views
165
• Calculus and Beyond Homework Help
Replies
1
Views
272
• Calculus and Beyond Homework Help
Replies
5
Views
315
• Calculus and Beyond Homework Help
Replies
6
Views
786
• Calculus and Beyond Homework Help
Replies
1
Views
903
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
2K
• Calculus and Beyond Homework Help
Replies
9
Views
1K