Line tangent to an elipse given one point

In summary: the gist of it is that you can find the point of contact by drawing a triangle with a right angle that has a base of x_1/a and height y_1/b, and then finding the angle between the base and hypotenuse, and the angle between the hypotenuse and the line joining the center of the circle and the contact points.
  • #1
thursdaytbs
53
0
Write the equations of the tangent lines from (3,0) to x^2 + 2y^2 = 6

Anyone have any idea? I've graphed it but i still have no clue where it's tangent to the elipse.
 
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  • #2
thursday,

What do you know about a curve at the point where a line is tangent to it?
 
  • #3
Okay, I'm in Precalculus Honors, and have not learned about Derivatives yet, and this is an extra credit problem.

I've gathered that you must find the derative of the curve at the point of tangency, and that is the slope of the line. I'll google some more, but any help?
 
  • #4
thrusday,

Well...if you were given this problem before you've learned how to differentitate then there must be some other way to do it. Of course you can find the slope of a line that's tangent to a circle without calculus, so maybe you can with an ellipse too. Although it'll probably be a lot messier!

I'm not sure that this works, but if you draw lines from the two foci of the ellipse to a point on the ellipse it looks as though the line that bisects the angle they make at the point might be perpendicular to the tangent line. Do you know whether that's true? If not, can you figure out whether it's true?
 
  • #5
Can you find any easy to detect differences between:

A line not intersecting the ellipse
A line tangent to the ellipse
A line intersecting the ellipse, but not tangent to it

?
 
  • #6
The ditto the teacher handed out to us was not his. A previous teacher had made it, and people were asking about it today, and for him to help us with it. He put it on the board, looked at it, and said - forget it, omit it. although he said he may give extra points for it solved.

Hurkyl - I don't understand what you're asking. If I'm able to detect the differences between those three situations? I suppose I can, by just graphing and seeing?
 
  • #7
That's one way to do it. I'm hoping you can find a less subjective method, though.
 
  • #8
Given a formula for an ellipse, [tex]{x^2\over a^2} + {y^2 \over b^2} = 1[/tex] and a point [tex]Q(x_1,y_1)[/tex] outside of the ellipse, to find the point of contact [tex]P(x, y)[/tex] on the ellipse:

There's an intuitive solution and a solution using differentiation.

I'll show you the differentiation version and then show you it is equivalent to the intuitive one.

Here goes the differentiation version:

let [tex]u={x\over a}; v={y\over b}[/tex],
then [tex]u^2 + v^2 = 1[/tex] and it is a formula for a circle of unit radius.

You can parametrize this and say that
[tex]u = x/a = \cos t[/tex], and
[tex]v = y/b = \sin t[/tex].

and you get [tex]x = a \cos t[/tex] and [tex]y = b \sin t[/tex].
Thus [tex]P=(a \cos t, b \sin t)[/tex]
The vector along the tangential line at P is parallel to [tex]{dP\over dt} = (-a \sin t, b \cos t)[/tex].
Thus the cross product [tex](Q-P) \times {dP\over dt} = 0[/tex]
Thus
[tex](x_1-a \cos t) b \cos t + (y_1- b \sin t) a \sin t = 0[/tex].

Now divide the both sides by [tex]ab[/tex] and you get
[tex](x_1/a - \cos t) \cos t + (y_1/b - \sin t) \sin t = 0[/tex],

or

[tex]{x_1\over a} \cos t + {y_1\over b} \sin t = 1[/tex], since [tex]\cos^2 t+\sin^2 t = 1[/tex]

Now define D as
[tex]D^2 = u_1^2 + v_1^2[/tex], where [tex]u_1 = x_1/a[/tex] and [tex]v_1 = y_1/b[/tex]
and also define
[tex]\tan\theta = v_1/u_1 ; \cos\theta = u_1/D ; \sin\theta = v_1/D[/tex]

and also define [tex]\cos\phi = 1/D[/tex]

Then the equation is
[tex]u_1 \cos t + v_1 \sin t = 1[/tex],
or
[tex]D (\cos\theta \cos t + \sin\theta \sin t) = 1[/tex],
or [tex]\cos(\theta-t) = 1/D = \cos\phi[/tex].

so you get

[tex]\theta - t = \pm\phi[/tex],

or [tex]t = \theta \pm\phi[/tex], where [tex]\theta = \arctan(1/D)[/tex] and [tex]\phi = \arccos(v_1/u_1)[/tex].
and [tex]D=sqrt(u_1^2 + v_1^2)[/tex], and [tex]u_1=x_1/a, v_1=y_1/b[/tex], where [tex]{x_1}[/tex] and [tex]{y_1}[/tex] are coordinates of Q.
then P is given as [tex]P = (a \cos (\theta\pm\phi), b \sin (\theta\pm\phi))[/tex].


In other words to get the parameter t where the line contacts P
draw a triangle with a right angle that has a base of [tex]x_1/a[/tex] and height [tex]y_1/b[/tex].
Then the hypotenuse is D(>1 if Q is outside of the ellipse).
Draw a circle of unit radius with the lower end of the hypotenuse as its center.
cast a tangent line from the upper end of the hypotenuse to the circle.
The angle between the base of the triangle and the hypotenuse is [tex]\theta[/tex],
and the angles between the hypotenuse and the line joining the center of the circle and the contact points
are [tex]\pm\phi[/tex] each.
and t is anyone of the sum of angles say [tex]\theta+\phi[/tex] or [tex]\theta-\phi[/tex].

So you can see it is equivalent to the figure that you will get if you shrinked the ellipse by [tex]a[/tex] horizontally
and by [tex]b[/tex] vertically, and the coordinates of the Q accordingly, into say [tex]Q'(x_1/a, y_1/b)[/tex],
then drew the tangent lines from Q' to the shrunk ellipse, now a unit circle.

The complicated mathematical operations I have performed up above is just to show the validity of this intuitional operation. so you know what to do now:

1. Shrink the coordinate system uniformly until the ellipse becomes a unit circle.
2. draw the tangent line from Q' to the resulting unit circle and get the contact points P'.
3. stretch the coordinate syst\em back to the original one(P' --> P).
 
Last edited:
  • #9
you can use limit as well. that would be using the derivative but you would derive the function yourself...
 
  • #10
You can first find the general equation of a line passing through the given point.

That is, substitute the point (3,0) in y=mx + c. You will get a relation between 'c' and 'm'. Substitue for 'c' in terms of 'm' to get the general equation of a line passing through (3,0)

Now solve this general equation and the equation of the ellipse. You will get two values of 'x' [Can you see how you get it?] in terms of 'm'.

Can you then find the value of 'm' for which the line is a tangent?

(Hint: The two values of 'x' you get are the points of intersection between the line and the ellipse).
 

Related to Line tangent to an elipse given one point

1. What is a line tangent to an ellipse?

A line tangent to an ellipse is a line that touches the ellipse at exactly one point, known as the point of tangency. This line is perpendicular to the ellipse's curve at that point.

2. How do you find the equation of a line tangent to an ellipse?

To find the equation of a line tangent to an ellipse, you need to have the coordinates of the point of tangency. From there, you can use the slope of the tangent line, which is equal to the negative reciprocal of the slope of the ellipse at that point. Finally, you can use the point-slope form of a line to write the equation of the tangent line.

3. Can a line be tangent to an ellipse at more than one point?

No, a line can only be tangent to an ellipse at one point. This is because a tangent line is defined as being perpendicular to the ellipse's curve at a single point.

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

Finding a line tangent to an ellipse is useful in many applications, such as physics and engineering. It allows us to determine the instantaneous rate of change of a variable at a specific point on the ellipse, which can help us make predictions and solve problems.

5. What are some methods for finding a line tangent to an ellipse?

There are a few methods for finding a line tangent to an ellipse, including using the point-slope form of a line, using the derivative of the ellipse's equation, or using the parametric equations of the ellipse. Some methods may be more suitable for certain situations, so it is important to understand and be familiar with multiple techniques.

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