Tangent line to ellipse

Therefore, the two points on the ellipse having a tangent line passing through the point (0,3) are (2.236, 4/3) and (-2.236, 4/3).
  • #1
bajazu
2
0
Given equation x^2/9 + y^2/4 = 1. Determine the two points on the ellipse having a tangent line passing through the point (0,3). Cant seem to figure this one out? I have found the derivative of the slope which is -4x/9y. I don't know how to use that slope in order to find a tangent line that passes through the points (0,3), which is not located on the orignal ellipse?
 
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  • #2
bajazu said:
I don't know how to use that slope in order to find a tangent line that passes through the points (0,3), which is not located on the orignal ellipse?

Once you have the general expression for the slope of the tangent line to a point (x,y) on the ellipse, you need to look at the slope of a line that runs from (x,y) to (0,3). That will be

m = (y - 3)/(x - 0) ,

which you now equate with the slope for a tangent line (in other words, having found the slope for a line running from any point on the ellipse through (0,3), we now want to find the ones that are tangent lines).

The expression you get from equating these slopes can then be combined with the original equation for the ellipse (by whatever method you find convenient) to solve for either x or y, after which you can then find the remaining coordinate. (If you solve for y first, you'll get a single value, which marks two points on the ellipse.)
 
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  • #3
dynamicsolo said:
Once you have the general expression for the slope of the tangent line to a point (x,y) on the ellipse, you need to look at the slope of a line that runs from (x,y) to (0,3). That will be

m = (y - 3)/(x - 0) ,




So i use this equation with my slope (y - 3)/(x - 0) = -4x/9y , i don't understand what i am suppose to put for the (x,y) would i use the point given even though it is not located on the ellipse. Once done with the equation i would set one variable equal to x and then plug the equation into the original problem?
 
  • #4
bajazu said:
So i use this equation with my slope (y - 3)/(x - 0) = -4x/9y , i don't understand what i am suppose to put for the (x,y) would i use the point given even though it is not located on the ellipse. Once done with the equation i would set one variable equal to x and then plug the equation into the original problem?

What the problem is asking for is to find tangent lines which pass through the external point (0,3) and which are tangent to the given ellipse. The values (x,y) we are solving for are points on the ellipse, so those coordinates must satisfy the equation for the ellipse.

The equation which matches the slopes gives us

(-4x) · x = (9y) · (y-3) , or

[tex]-4x^{2} = 9y^{2} - 27y[/tex].

The solutions we are looking for must satisfy both this equation, so that the lines were are dealing with will be tangent to the ellipse, and the equation for the ellipse, since the tangent points are on the ellipse. If you solve this last equation for [tex]x^{2}[/tex], you can substitute that into the ellipse equation to solve it for y (you should get two values). Then use either equation to solve for the x-coordinates.
 
  • #5
bajazu said:
Given equation x^2/9 + y^2/4 = 1. Determine the two points on the ellipse having a tangent line passing through the point (0,3). Cant seem to figure this one out? I have found the derivative of the slope which is -4x/9y. I don't know how to use that slope in order to find a tangent line that passes through the points (0,3), which is not located on the orignal ellipse?


The equation of a tangent line to a given point of the ellipse (x1,y1) is x.x1^2/a^2+ y.y1^2/b^2=1. For the lines that passes from the point(0,3) and are tangent to the ellipse we have 0.x1/9+3.y1/4=1, then to get the coordinats of the points where the tangent lines touch the ellipse we have to solve the system of two equations: 0.x1/9+3.y1/4=1 and x1^2/9+y1^2/4=1, and we get the poins(2.236,4/3);(-2.236,4/3)(2.236 or square root of 5)
 

Related to Tangent line to ellipse

What is a tangent line to an ellipse?

A tangent line to an ellipse is a line that touches the ellipse at exactly one point, without intersecting the ellipse at any other point. This point of contact is known as the point of tangency.

How is the equation for a tangent line to an ellipse calculated?

The equation for a tangent line to an ellipse can be calculated using the formula y = mx + b, where m is the slope of the tangent line and b is the y-intercept. The slope of the tangent line can be found by taking the derivative of the ellipse equation at the point of tangency.

What is the relationship between the slope of a tangent line and the slope of the ellipse at the point of tangency?

The slope of the tangent line to an ellipse is perpendicular to the slope of the ellipse at the point of tangency. This means that the product of the two slopes is equal to -1.

Can there be more than one tangent line to an ellipse at a given point?

No, there can only be one tangent line to an ellipse at a given point. This is because a tangent line must touch the ellipse at exactly one point, and any other line passing through that point would intersect the ellipse at another point.

What is the significance of finding the tangent line to an ellipse?

Finding the tangent line to an ellipse is important in various fields of science and engineering, as it allows for the determination of the rate of change (or slope) at a specific point on the ellipse. This information can be used for optimization, curve fitting, and other applications.

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