calc question, my teacher couldnt even do it?

Question

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)

Answer 1

1. Find dy/dx by implicit differentiation

dy/dx = -(b^2x) / (a^2y)

2. Form eqn of tangent at point say (p,q) on ellipse.

y -q =-(b^2p) / (a^2q)(x-p)

3.Find intercepts of this line with axes

y int y=(p^2b^2)/(a^2q) +q = b^2/q

x int x = (p^2b^2 +a^2q^2)/(b^2p) = a^2/p


4. Form eqn for area of triangle

A = (a^2b^2)/(pq)

Write in terms of 1 variable
A=(a^2b^2) / (p sqr((a^2b^2-b^2p^2)/a^2))
A= a^3 b p^-1 (a^2 -p^2) ^ (-1/2)

5. Find when A' =0

A' = a^3b/p -1/2 ( a^2-p^2)^(-3/2) (-2p) +(a^2 -p^2)^(-1/2). -a^3b/p^2

= a^3b { p^2 -(a^2-p^2)}/ denominator

= 0 if p^2 =a^2 /2
p=a/sqr2

p=a/sqr2 =.> q = b/sqr2

6. Sub into eqn of tangent

y - b/sqr2 = (-b/a) (x- a/sqr2)

y = (-b/a) x + 2b/sqr2

Answer 2

Since we are dealing with the first quadrant, rewrite the equation:
b^2x^2 + a^2y^2 = a^2b^2 becomes

y = sqrt[(-b^2/a^2)x^2+b^2] (I'll leave those details to you). x is limited to the range [0,a].

y' = (-b^2/a^2)(x)*[(-b^2/a^2)x^2+b^2]^(-1/2) (again, I leave the details to you). This will give us the slope of the tangent line at a point.

Pick a point k in [0,a]. The actual point will be (k, sqrt[(-b^2/a^2)k^2+b^2]) and will have slope (-b^2/a^2)k*[(-b^2/a^2)k^2+b^2]^(-1/2).

We can use the Point-Slope form to find the equation of the tangent line at the point x=k. I am going to substitute v = sqrt[(-b^2/a^2)k^2+b^2]

y - v = (-b^2/a^2)(k/v)*(x-k) which finally gives us
y = ((-b^2k)/(a^2v))x + (1+(b^2k^2/a^2v))

From this, when x=0, y=(1+(b^2k^2/a^2v))
and when y=0, x=[1+(b^2k^2/a^2v)]/[(b^2k)/(a^2v)].

The area of the triangle formed by the tangent line will be 1/2 * value of y-intercept * value of x-intercept.

*whew* That is incredibly messy, but accurate.

An educated guess would tell us that the minimal triange will occur at an angle of 45 degrees, so find the point on the ellipse at that angle and go from there.

Answer 3

Using polar co-ordinates

x=a cos p
y=b sin p

Slope of tangent dy/dx = -b/a tan p

Equation of tangent

y - b sin p = -b/a tan p (x - a cos p) .... Eqn 1

in Eqn 1

set x = 0 yields y = b/ sin p

and

set y=0 yields y = a/ cos p

Area of triangle = 0.5ab / sin p cos p = ab / sin (2p)

Differentiating dA/dp = -2ab cos (2p) / sin^2 (2p)

dA/dp = 0 when p = pi/4 for 0<=p<=pi/2

Thus x = a / 2^0.5 and y = b / 2^0.5

Substituting into Eqn 1 will yield

y = 2^0.5b - b/a x or bx + ay = 2^0.5 ab

Note The minimum area is ab

Answer 4

Let (x0,y0) the point of tangence in the ellipsis and (0,0), (X,0) and (0,Y) the coordinates of the vertices of the triangle.

Derivating implicitly the equation above, we have:

2b²(x0) + 2a²(y0)(dy/dx)(0) = 0, where (dy/dx)(0) denotes the derivative in the point (x0,y0).

Then we have:

dy/dx(0) = -b²(x0)/a²(y0) (I)

The equation of the line connecting (X,0) and (0,Y) is:

yX + xY = XY. (II)

Observe that tan(arc line) = -Y/X = (dy/dx)(0) and by (I) we have:

Y/X = b²(x0)/a²(y0) (III)

Note that (x0,y0) satisfies (II). Therefore, using (II) and substituting (III) we have :

y0 = -b²(x0)/a²(y0) + Y or

Y = b²/y0

In a similar way

X = a²/x0

The area of the triangle is:

S = XY/2 = 1/2(a²b²/x0*y0) and as (x0,y0) belongs to the ellipsis
we have that

S = 1/2(a³b/x0*√a²-x0²) which is minimum for x0 = (a√2)/2
Therefore y0 = (b√2)/2.

This way we are able to calculate X and Y:

X = a√2 and Y = b√2 and the equation of the line is:

ay + bx = √2ab.