Line tangent to an elipse given one point

[thursdaytbs]

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.

 

[jdavel]

thursday,

What do you know about a curve at the point where a line is tangent to it?

 

[thursdaytbs]

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?

 

[jdavel]

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?

 

[Hurkyl]

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

?

 

[thursdaytbs]

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?

 

[Hurkyl]

That's one way to do it. I'm hoping you can find a less subjective method, though.

 

[mingshey]

Given a formula for an ellipse, {x^2\over a^2} + {y^2 \over b^2} = 1 and a point Q(x_1,y_1) outside of the ellipse, to find the point of contact P(x, y) 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 u={x\over a}; v={y\over b},
then u^2 + v^2 = 1 and it is a formula for a circle of unit radius.

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

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

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

or

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

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

and also define \cos\phi = 1/D

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

so you get

\theta - t = \pm\phi,

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


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 x_1/a and height y_1/b.
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 \theta,
and the angles between the hypotenuse and the line joining the center of the circle and the contact points
are \pm\phi each.
and t is anyone of the sum of angles say \theta+\phi or \theta-\phi.

So you can see it is equivalent to the figure that you will get if you shrinked the ellipse by a horizontally
and by b vertically, and the coordinates of the Q accordingly, into say Q'(x_1/a, y_1/b),
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).

 

[wisredz]

you can use limit as well. that would be using the derivative but you would derive the function yourself...

 

[siddharth]

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).