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### Calculus Problem

```
Date: Tue, 29 Nov 94 21:56:34 CST
Subject: Problem

I'm a senior in a college calculus class.  The following problem
was assigned, but the next day, after no one had completed it,
our professor attempted to do it but quit after 25 minutes.

Find the equation of the line tangent to the ellipse
b^2*x^2   +    a^2*y^2   =   a^2*b^2
in the first quadrant that forms with the coordinate axes
the triangle of smallest possible area
(a & b are positive constants)

Thanks!
```

```
Date: 7 Dec 1994 00:46:14 GMT
From: Dr. Ken
Subject: Re: Problem

Hello there!

I'll give this problem a shot, and see whether I can think of a good
way to do it.

First, we'll try to find the equation of a line that's tangent to the
ellipse.  Since you said that we're looking for tangency in the first
quadrant, I'll write the equation of the ellipse as

y = b*Sqrt{a^2 - x^2}/a

I've just solved for y here.  Then to find the slope of the tangent line
to the ellipse, I'll take the derivative of this expression:

y' = -bx/(a*Sqrt{a^2 - x^2})

So if we let r be some value of x that gives us a point on the ellipse in
the first quadrant, the equation of the line tangent to the ellipse at
(r,b*Sqrt{1-(1/a)^2*r^2} will have the form

y = (-br/(a*Sqrt{a^2 - r^2})) x  + c.

The next step is to find c in terms of a, b, and r.  So let's do it. The
usual way is to plug in (for x and y) some point that you know is on the
line, and in this case we'll pick the point (r, b*Sqrt{a^2 - r^2}/a). So
we get

b*Sqrt{a^2 - r^2}/a = -br^2/(a*Sqrt{a^2 - r^2}) + c

c = b*Sqrt{a^2 - r^2}/a + br^2/(a*Sqrt{a^2 - r^2})

b(a^2 - r^2)            b*r^2                 ab
c = ----------------- + -----------------  =  ---------------
a*Sqrt{a^2 - r^2}   a*Sqrt{a^2 - r^2}     Sqrt{a^2 - r^2}

So the equation for the line tangent to the ellipse at the point
(r,b*Sqrt{a^2-r^2}/a) is

-brx                 ab               a^2*b - brx
y = -----------------  + ---------------  =  -----------------
a*Sqrt{a^2 - r^2}    Sqrt{a^2 - r^2}     a*Sqrt{a^2 - r^2}

Now take a breath.  We've completed the first part of the problem.
Go get a glass of orange juice or something, because there's more,
and we're going to start to use Calculus.

Okay, we're back.  We want to find out the area of the region bounded by
this line and the x- and y- axes.  Well, it's a right triangle, so we'll
use the equation Area = Base x Height / 2.  The base and the height are
going to be, respectively, the x-intercept and the y-intercept of the
line we found.  So let's find the intercepts:

To find the y-intercept, we plug in x = 0 and solve for y, finding
y = ab/Sqrt{a^2 - r^2}.  Similarly, to find the x-intercept, we plug in
y = 0 and solve for x, finding that x = a^2 / r.  So the area of the
triangle is going to be the product of these numbers divided by 2, i.e.
a^3*b/(2r*Sqrt{a^2 - r^2}.

Take another breath.  A smaller breath, though, because that part wasn't
as hard.  Now we need to treat this like an ordinary Max/Min problem,
and find the minimum value of this Area function.  The way we do this
is to take its derivative, and then set that equal to zero to find
critical points.  So let's do it.  Remember, the area is a function of
only one variable, since a and b are fixed; they're given to us in the
beginning of the problem.

-a^3*b (Sqrt{a^2 - r^2} - r^2 / Sqrt{a^2 - r^2})
Area' = ------------------------------------------------
2r^2 (a^2 - r^2)

Check my work here.  Not only is it important that you see how the
computations work, I might have messed up.

So we want to set this equal to zero and solve for r.  Remember that a
fraction is zero when its numerator is zero, so we solve the equation:

-a^3*b(Sqrt{a^2 - r^2} - r^2/Sqrt{a^2 - r^2}) = 0.

a^5*b
I get r^2 = ---------
a^3*b - 1

Wow, that's a nice problem. I've certainly never seen it before, but
I'll keep it in mind whenever people bug me for a problem.

If you couldn't follow something, please write me back and I'll see

-K
```

```
Date: 6 Jun 1995 12:04:19 -0400
From: Dr. Ken
Subject: Re: Correction!

>Just found the Dr. Math archives.  I was interested in the problem of
>finding the equation of the tangent line to an ellipse that formed the
>smallest possible triangle with the coordinate axis in the first
>quadrant (reference in subject line).  I didn't take time to check your
>work, but I did it with PARAMETRIC EQUATIONS, and got a different

NB. Bob Dillon's solution is given below.

>Assuming the calculus class had studied parametric equations, this
>seems more elegant (and simpler).
>

Hello there -

Thanks for the message.

Your work looks good, and I agree that it's a nice solution.  Thanks
for sending it in!

-Ken
```

```
BOB DILLON'S SOLUTION

Problem: Find the equation of the tangent line to the ellipse
x^2/a^2 + y^2/b^2 = 1  that cuts off the smallest area with the
coordinate axes in the first quadrant.

The ellipse has parametric equations x = a cos t   y = b sin t

dy     b cos t
which gives  x' = -a sin t   y' = b cos t  and   -- = - -------
dx     a sin t

Then the equations of the tangent line at (a cos t, b sin t) are
x = a cos t + (a sin t) u,  y = b sin t - (b cos t) u  [u a parameter]
and this gives x=0 when u = - cot t,  y=0 when u = tan t,
which in turn gives intercepts

x_0 = a cos t + a sin t tan t = a(cos t + sin^2 t/cos t)

= a(cos^2 t + sin^2 t)/cos t = a/cos t

y_0 = b sin t + b cos t cot t = b(sin t + cos^2 t/sin t)

= b(sin^2 t+ cos^2 t)/sin t = b/sin t

Then the area of the triangle is (1/2) x_0 y_0
= ab/(2 sin t cos t) = ab/sin 2t = ab csc 2t

and dA/dt = -2ab csc 2t cot 2t = 0  when t=pi/4

This gives the equations of the tangent line as
x = a/sqr(2) (cos t + u sin t)  y = b/sqr(2) (sin t - u cos t)
or for non-parametric form, we have (x, y)=(ka, kb), with slope (-b/a)
so y-kb=(-b/a)(x-ka)  or y = (-b/a)x + 2kb  or bx+ay=2kab
where k=1/sqr(2) = sin (pi/4) = cos (pi/4)

Note that the area of the triangle is ab.

Incidentally, the answer for t is the same as for a circle.

```

```
Date: 06/14/2001 at 03:17:14
From: Doraci Gabriel da Rosa
Subject: Calculus

Not a question, but a specific comment. I found the same solution
the other way.

b^2*x^2 + a^2*y^2 = a^2*b^2 ==> 2(b^2*x + a^2*y*y') = 0 ==> y'
= -(b/a)^2*(x/y)

Then the equations of the tangent line at (x_0, y_0) are

(...) ==> b^2*x_0*x + a^2*y_0*y = a^2*y_0^2 + b^2*x_0^2, but
a^2*y_0^2 + b^2*x_0^2 = a^2*b^2, so

b^2*x_0*x + a^2*y_0*y = a^2*b^2 ==> x/(a^2/x_0) + y/(b^2/y_0) = 1

Then (a^2/x_0)  e (b^2/y_0) are the base and height of rectangle
triangle.

So, the area of the triangle is

A = a^2*b^2/(2*x_0*y_0)

dA/dx_0 = 0 ==> a^2*b^2*(- y_0 - x_0*y_0')/(2*x_02*^y_0^2) = 0 ==>
(...)

Solution:

x_0 = sqrt(1/2)*a ; y_0 = sqrt(1/2)*b

Hence, the equation of the tangent line is

b*x + a*y - sqrt(2)*a*b = 0  ( ! )
```

```
Date: 04/17/2003 at 14:26:00
From: Ray Van Raamsdonk

I thought this was an interesting problem, and you might be interested
in seeing a couple of additional solutions.

My first attempt at this was like anyone else's approach using messy
equations, differentiating expressions containing square roots and the
like. However the problem is rather nice and so the solution should
also be. So this was my second approach:

1. X^2/a^2 + Y^2/b^2 = 1   Equation for an Ellipse

2. X*A/a^2 + Y*B/b^2 = 1   Equation for a Tangent to the Ellipse
passing through coordinates (A,B)

3. From 2. the tangent intersects the X axis at X= a^2/A and
the Y axis at Y = b^2/B

4. From 3. the area of the triangle is AREA = 1/2*(a^2/A)*(b^2/B)

5. This area is a minimum when A*B is a maximum.

6. From 1. B^2 = (b^2)*(1-A^2/a^2)

7. Maximizing (A^2)*(B^2) also maximizes A*B  (avoids square roots)

8. So (A^2)*(B^2) = (A^2)*(b^2) - (A^4)*(b^2)/(a^2)

9. The maximum of this expression which can be found using the
completion of squares method or calculus occurs when:

A = a/sqrt(2) and B = a/sqrt(2)

10. Then the area of the square is a*b

11. The equation for the tangent line is:

X/(a*sqrt(2)) + Y/(b*sqrt(2)) = 1

An even shorter solution by inspection use the following idea by my
son Mark Van Raamsdonk:

It's easiest to just take the answer for a circle of radius 1,

x + y = sqrt(2)

and then map this problem to the more general one by a linear
transformation x -> x/a, y -> y/b which preserves ratios of areas.

-Ray

```
Associated Topics:
College Calculus

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