Conceptualizing two tangent lines to an ellipse

AI Thread Summary
An ellipse has only one tangent line at any given point on its curve, but a point outside the ellipse can have two tangent lines connecting to it. The discussion clarifies that the original problem involves finding the equations of two tangent lines from the point (12, 3) to the ellipse defined by x^2 + 4y^2 = 36. Participants emphasize the importance of visualizing the problem by graphing the ellipse to understand the tangent lines better. The conversation also touches on the complexities of deriving tangent equations, suggesting the use of established formulas for clarity. Ultimately, the key takeaway is that while a point on the ellipse has one tangent, external points can connect to the ellipse with two tangents.
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I'm assuming that given point can only have one tangent line because it's just the instantaneous slope at a point. If so, then how can an ellipse have two tangent lines at a point? Do they mean something else?
 
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Does WHO mean something else? Since you haven't told us what the problem says, we can't possibly tell what they might mean! Yes, an ellipse, like any smooth curve has only one tangent line at any point on the ellipse. But it is conceivable that the problem you are thinking of is asking about two tangent lines from some point that is not on the ellipse.
 
Perhaps they mean two tangents in opposite directions?
 
It can't.A tangent line is a LINE (it's infinite).It comes from -\infty and goes to +\infty...

Daniel.
 
I meant vectors...
 
The OP meant "tangent lines"...

Daniel.
 
Clarification: "Find the equations of both the tangent lines to the ellipse x^2+4y^2 = 36 that pass through the point (12,3)
 
That's something else.Plot the ellipse and u'll see that one of the tangents is y=3 and the other can be found from

\frac{x^{2}}{6^{2}}+\frac{y^{2}}{3^{2}}=1 (the ellipse)

\frac{x-x_{0}}{6^{2}}+\frac{y-y_{0}}{3^{2}}=1
(the tangent to ellipse passing through the point \left(x_{0},y_{0}\right))


Daniel.
 
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In case it wasn't clear from dexter's response, the answer to your question is that there is only one tangent to an ellipse at a given point on the ellipse, but the point in that problem is well off the ellipse (try plugging it into the equation and you'll see that it doesn't work). To actually answer the question, you can use the formulae dexter gave you.
 
  • #10
ohhhhh So that point is floating off in space and, seeing that, of course there two tangent lines that can run though it on the ellipse. I should've graphed it to better visualize. Thank you both.
 
  • #11
Always plot the graph of the function.The point was away from the ellipse and from it,u could build two tangents...

Daniel.
 
  • #12
Why does it seem so complex? Is it not possible to find the equation of a line from the origin of the ellipse (assumed (0,0)) and then find the perpendicular gradient and then the equation of the tangent or have I missed something that would not make this a feasible method?

The Bob (2004 ©)
 
  • #13
I didn't understand a thing :confused: What gradient?

Daniel.
 
  • #14
I believe my approach is trial and error but from the origin to a point on the ellipse find the gradient and then the perpendicular to that which pass the point (12,3).

I tell you what though, it sounds more stupid the more I think about it.

What do you have to do then?

The Bob (2004 ©)
 
  • #15
Apply the HS taught formula giving the equation of a tangent to an ellipse...?(v.post #8).

Daniel.
 
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