Calculus Question - Tangents to an ellipse - its got me stumped

[zacl79]

Calculus Question - Tangents to an ellipse - its got me stumped!

Homework Statement

A tangent line is drawn to the ellipse x^2/25 + y^2/16 = 1 so that the part intercepted by the co-ordinate axis is a minimum. Show that it has a length of 9 units.

Homework Equations

x^2/25 + y^2/16 = 1

The Attempt at a Solution

I have seen this question on the forums, through a quick google search, but cannot see how the user came up with the steps and eventual answer. I think you need to differentiate the equation first and this will give you the slope at the point (where the intercepted part is a minimum). From there i can only guess, something like constructing an equation for tangent, and then solving for the x and y intercepts. Substitute into pythagoras.
I have given this a shot, but to no avail. Please Help!

 

[vela]

What does "part intercepted by the co-ordinate axis" mean?

 

[D H]

It means "part intercepted by the coordinate axes", not axis. Terribly worded problem, particularly when misspelled.

All tangent lines except for the purely vertical and horizontal ones will intercept both the x- and y- axes. The origin and the points of intersection with the two axes will thus form a triangle. It is the length of the hypotenuse of this triangle that is to be minimized.

And that explanation also suffices as a hint.

 

[zacl79]

Sorry i MEANT axes, and i understand that they will form a triangle and hence minimize the triangleto find the minimum length. It just a matter of how i go about doing it, is what i was asking for help with.

 

[vela]

I figured you meant "axes," but I never heard the phrase "part intercepted" before. I interpreted it as the x and y intercepts, not the line segment between those two points.

Anyway, you said you made an attempt. Why don't you explain what you did, with details, so we can try figure out where the problem is.

 

[zacl79]

Yeah, i believe that your interpretation is correct, the questions wants the length of the hypotenuse (the distance between the y intercept and the x intercept).

Firstly i differentiated the x^2/25 + y^2/16 = 1 =
f'=-16x/25y the slope of the tangent where it meets the curve.

I believe that you need to construct and equation for tangent in the form of
(y-yo)=m(x-xo).
And once that is constructed solve for y when x=0 and solve for x when y=0. This gives the two sides to a pythagoras triangle. Solving for the hypotenuse.

Now that was what i thoguht you had to do, i had a problem in getting the point (x,y) where the tangent meets the curve.
An idea i had was to find the minimum rectangle that could fit inside the ellipse, and the vertex would meet the point where the tangent does.
It went something like this:
A=4xy
A^2=256x^2 - 256x^4/25 (substituting for y)
I think you diffrentiate to get the values fro x and y, but that is as far as i got, I am not sure if I am doing the right thing, or if my maths is correct.

Thanks

 

[vela]

For a given x₀, you can find the corresponding y₀ using the equation of the ellipse. Or you could try using the parametrization x₀=5 cos θ, y₀=4 sin θ, and write everything in terms of θ.

 

[zacl79]

But in order for me to get the points where the tangent meets the axes, don't i need to know the equation of the tangent?
As for the parametrization, i haven't done it like that in a few years, ill have a go later if the way I am doing it doesn't work.

 

[vela]

When I wrote x₀ and y₀, I was not referring to the x and y intercepts. I was referring to the quantities in the equation you wrote for the line, y-y₀=m(x-x₀).

 

[zacl79]

oh i see, I am sorry that was a misunderstanding. I am aware that i can get x0 from y0 and vice versa. But its a matter of getting at least one point first, was i on the right track with the rectangle idea?

 

[vela]

No, if you try to minimize the area, you'll end up with points on the axes; in other words, rectangles with zero area. You could try maximizing the area, but I don't see how it follows then that the line tangent to that point meets the problem's criterion of having minimal length between the axes.

What I'm suggesting is you start with a point on the ellipse and find the equation of the line that's tangent to the ellipse at that point.

 

[zacl79]

the way i interpretted it was that the point where the tangent touches of the minimal length will be the same poit as if i was to maximise the rectangle. Obviously not then.

Im sorry if I am not picking up on something trivial, by taking a random point on the ellipse and construcing an equation for the tangent, how is that meant to get me any closer to the tangent at the point where minimal length occurs?

Thanks

Thanks

 

[rs1n]

You started with the right idea when computing the formula for the slope of the hypotenuse, as well as using the point-slope form of a line. Now suppose the point of tangency occurs at (a,b). Then the slope is

$$m=-\frac{16a}{25b}$$

and the equation of your tangent line is

$$y-b = -\frac{16a}{25b}(x-a)$$

From here, you can compute the coordinates of the intercepts in terms of a's and b's, and then use the distance formula to find the length of the hypotenuse. In the end, you will have a formula for the length in terms of a's and b's. Let's call this formula L(a,b). You need to minimize L(a,b). This can be done several ways, depending on how much calculus you know. Note that L(a,b) is just distance formula and has a square root -- you can actually just consider the square of the length (think about why this is true) to avoid having to deal with square roots when finding extrema.

If you know about derivatives, then use the fact that (a,b) is a point on both the line and the ellipse, so that (a,b) satisfies

$$\frac{a^2}{25}+\frac{b^2}{16}=1$$

This formula may help you transform L(a,b) into a function of just a's or just b's... then find the minimum the usual way.

Another method (if you have had multivariables) is to use the technique of optimization via Lagrange multipliers to finish the problem.

 

[zacl79]

okay as far as i got was setting up two equations for the coordinates at the x and y axes.

y=-16a^2/25b -b
x= 25b^2/16a + a

Are you saying these now go into the L = x^2 + y^2

sqr((625b^4/256a^2 + 50ab^2/16 + a^2) + (256a^4/625b^2 -32ba^2/25 + b^2))
I know that is difficult to comprehend, but is that what it sort of got to look like?

Then following all that differentiate to find the minimum.

 

[vela]

the way i interpretted it was that the point where the tangent touches of the minimal length will be the same poit as if i was to maximise the rectangle. Obviously not then.

Interpreted what? Your own idea? You seemingly pulled this idea out of thin air and asked if it was the right method. Do you have a reason to think that method would work? Was there a hint given?

 

[zacl79]

Sorry, the way i interpreted the question, is all i was saying. I didnt mean to say that my idea was right, i thought we were discussing the same idea, but i was wrong, I am sorry for any confusion.

 

[rs1n]

okay as far as i got was setting up two equations for the coordinates at the x and y axes.

y=-16a^2/25b -b
x= 25b^2/16a + a

Are you saying these now go into the L = x^2 + y^2

sqr((625b^4/256a^2 + 50ab^2/16 + a^2) + (256a^4/625b^2 -32ba^2/25 + b^2))
I know that is difficult to comprehend, but is that what it sort of got to look like?

Then following all that differentiate to find the minimum.

Almost. First, the intercepts are:

$$\left(0, \frac{16a^2}{25b}+b\right) \quad \text{and} \quad \left(\frac{25b^2}{16a}+a,0\right)$$

While the length of the hypotenuse is indeed

$$L=\sqrt{x^2+y^2}$$

(and this is what we need to minimize), you can actually just use

$$L=x^2+y^2$$

because the minimum of x^2+y^2 also produces the minimum of sqrt(x^2+y^2). Now, if you use the intercepts as-is to get a formula for L, you will get a mess. So let's try for some shortcuts. Notice that

$$\frac{a^2}{25}+\frac{b^2}{16}=1 \implies 16a^2 = 25\cdot (16 - b^2)$$

so that y-intercept is

$$\frac{16a^2}{25b} +b = \frac{25\cdot (16-b^2)}{25b}+b = 16b^{-1}-b +b = 16b^{-1}$$

Now show (through similar steps) that the x-intercept is equal to 25a^(-1). Putting these results together should produce:

$$L=x^2+y^2 = \left(\frac{25}{a}\right)^2+ \left(\frac{16}{b}\right)^2 = \frac{25^2}{a^2} + \frac{16^2}{b^2}$$.

Use the the fact that

$$\frac{a^2}{25}+\frac{b^2}{16}=1 \implies a^2 = 25\left(1-\frac{b^2}{16}\right)$$

and rewrite L in terms of just b. Then minimize.

 

[zacl79]

Thanks for that, ill give it a go tomorrow and see if that works. Just outta interest have you solved the problem to get the minimum distance to be 9 units??

 

[zacl79]

okay, i rearranged and everything and got L=(-375b^2 + 6400)/(25b^2 - 25b^4/16) and then minimized to get something looking like this:

(-320750/b -61250b - 3125b^3)/(25-25b^2/16) = 0

and I am not to sure how to solve it, that's if i minimized correctly!

 

[D H]

Almost. First, the intercepts are:

$$\left(0, \frac{16a^2}{25b}+b\right) \quad \text{and} \quad \left(\frac{25b^2}{16a}+a,0\right)$$

Before you go further with this, simplify this.

Denote the x- and y-intercepts as

$$\aligned x &= \frac{25b^2}{16a}+a \\ \\ y &= \frac{16a^2}{25b}+b \endaligned$$

or

$$\aligned x &= \frac{16 a^2 + 25b^2}{16a} \\ \\ y &= \frac{16a^2 + 25 b^2}{25b} \endaligned$$

That common numerator reduces to a simple number. Using the fact that

$$\Bigl(\frac a 5 \Bigr)^2 + \Bigl(\frac b 4 \Bigr)^2 = 1$$

Then

$$16 a^2 + 25 b^2 = 16*25$$

Thus

$$\aligned x &= 25/a \\ y &= 16/b \endaligned$$

You might find working with this a bit easier.

 

[zacl79]

so just to get this right, the x=25/a and y=16/b gets substituted into L = x^2 + y^2, isn't that just the same as i was doing before?? have you managed to get the answer to be 9 units??

 

[vela]

so just to get this right, the x=25/a and y=16/b gets substituted into L = x^2 + y^2, isn't that just the same as I was doing before?

Yes.

Gave you managed to get the answer to be 9 units?

Yes, the answer comes out to be 9.

 

[vela]

The equation of the ellipse gives you

$$\frac{a^2}{25}=1-\frac{b^2}{16} \Rightarrow \frac{25}{a^2} = \frac{1}{1-\frac{b^2}{16}}$$

and you found

$$L=\frac{25^2}{a^2}+\frac{16^2}{b^2}=\frac{25}{1-\frac{b^2}{16}}+\frac{16}{\frac{b^2}{16}}$$

To simplify the algebra, let z=b²/16. Then

$$L=\frac{25}{1-z}+\frac{16}{z}$$

Find the z that minimizes that function and plug it back into solve for L (which is the length squared, remember).

 

[zacl79]

thanks ill get on to it and get back to you how it goes. thanks

 

[zacl79]

Thanks very much, i know understand how to do a problem like this should it creep up again, and i worked it all out for length to = 9 units. thanks heaps everyone