Finding Tangent lines on an Ellipse that intersect through an external point

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zonova

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Hello, i'm having a rather massive problem. In fact, our entire class can't seem to figure out what to do. The problem is as follows:

"Find the equation of both tangent lines to the ellipse x^2 + 4y^2 = 36 that passes through the point (12,3)."

I don't quite no where to start. I've done problems where we had to find the equation of a tangent line at a point, but in this problem the point does not seem to be on the Ellipse. Any help on how to start would be great. I was thinking that i would start by solving for x, and then for y. However, that just gives me two equations. Perhaps i could set up a system with the two? Even then, i have no idea on where/how i would take that further. Thank you in advanced for any help you can give
 
What is the slope of ANY point on your ellipse?

Note: Implicit Differentiation makes this rather simple to calculate.

Note: Calculating the general derivative for any point may be relatively easy, but then you ust know what to do with it.

What's the slope of any line containing the given point and any point on the ellipse? If you have this, you should be able to write the equation of the general line, since you've already the given point.
 
tkhunny said:
What is the slope of ANY point on your ellipse?

Note: Implicit Differentiation makes this rather simple to calculate.

Note: Calculating the general derivative for any point may be relatively easy, but then you ust know what to do with it.

What's the slope of any line containing the given point and any point on the ellipse? If you have this, you should be able to write the equation of the general line, since you've already the given point.
Click to expand...



Ok, so using implicit differentiation, to find the slope of any point on this ellipse would be:
y' = -x/4y. Is that right?

Ok, so i understand what you are saying at the end. However, i'm having trouble figuring out how to find that slope.
 
Pick any point on the ellipse. Call it (a,b). Use this "point" and the given point to create an equation of a line.

Using the linear equation, and using your slope from the ellipse, what can we do?
 
Any line through (12, 3) can be written as y= m(x- 12)+ 3. Be sure to distinguish between the (x, y) as variables and \(\displaystyle (x_0, y_0)\) where the line is tangent to the ellipse. So you must have both \(\displaystyle x_0^2+ 4(m(x_0-12)+ 3)^2= 36\) and, as you say, \(\displaystyle m= -\frac{x_0}{4(m(x_0- 12)+ 3)}\), two equations to solve for \(\displaystyle x_0\) and m.
 
zonova said:
Hello, i'm having a rather massive problem. In fact, our entire class can't seem to figure out what to do. The problem is as follows:

"Find the equation of both tangent lines to the ellipse x^2 + 4y^2 = 36 that passes through the point (12,3)."

I don't quite no where to start. I've done problems where we had to find the equation of a tangent line at a point, but in this problem the point does not seem to be on the Ellipse. Any help on how to start would be great. I was thinking that i would start by solving for x, and then for y. However, that just gives me two equations. Perhaps i could set up a system with the two? Even then, i have no idea on where/how i would take that further. Thank you in advanced for any help you can give
Click to expand...

Step-1 find y' using eqn. of ellipse.
Step-2 consider any point (a,b)on ellipse and find tangent on that point
Step-3 In order to find tangent through (12,3).........(12,3) will lie on the tangent obtained in step-2
Step-4 solve the eqn. obtained in step-3 and you will get two values of x. corresponding to these values of x. find the values of y. in this way we will get two points on ellipse where the tangent through (12,3) meet ellipse.
Step-5 corresponding to points in step-4 you will get two tangents.
 
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