Tangent line to ellipse

[bajazu]

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?

 

[dynamicsolo]

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.)

 

[bajazu]

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?

 

[dynamicsolo]

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

$$-4x^{2} = 9y^{2} - 27y$$.

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 $$x^{2}$$, 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.

 

[Adem Meta]

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)