Calculus ProblemDate: Tue, 29 Nov 94 21:56:34 CST From: Brad Robert Plaster 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 if I can help you out. -K Date: 6 Jun 1995 12:04:19 -0400 From: Dr. Ken Subject: Re: Correction! As Bob Dillon <bdillon@admin.aurora.edu> wrote, >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 >answer (I think). NB. Bob Dillon's solution is given below. >Assuming the calculus class had studied parametric equations, this >seems more elegant (and simpler). > >Bob Dillon <bdillon@admin.aurora.edu> 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. Bob Dillon <bdillon@admin.aurora.edu> 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 Subject: Additional solutions 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. I hope these are of interest to your readers. -Ray |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/