Find the point on the curve y =2/3 〖√(18-x^2 )〗 (first quadrant) where a tangent line may be drawn so that the area of the triangle formed by the tangent line and the coordinate axes is a minimum.

Answer 1

we have

4x^2 + 9y^2 = 72
8x + 18y y' = 0
y' = -4x/9y
So, the line through (h,k) tangent to the curve is

y-k = (-4h/9k) (x-h)
That line has x-intercept at
(-4h/9k)x + 4h^2/9k + k = 0
x=(4h^2+9k^2)/(9k) * (9k/4h)
= (4h^2+9k^2)/(4h)
But, since 9k^2 = 72-4h^2, the x-intercept is at
(4h^2+72-4h^2)/(4h) = 1/(18h)

They-intercept is where x=0, or
y-k = (-4h/9k) (-h)
y = 4h^2/9k + k
= (4h^2+9k^2)/9k
= 1/(8k)
= 1/(8 * 2/3 √(18-h^2))
= 1/(12√(18-h^2))

The area of the triangle is 1/2 bh, or

a = (1/2)(1/18h)(1/(12√(18-h^2))
= 1/(432h√(18-h^2))

da/dh = (h^2-9)/(216h^2(18-h^2)^(3/2))
da/dh=0 when h=3.

So, the line through (3,2) defines the triangle with minimum area. That line has slope -2/3.

See the graphs at

http://www.wolframalpha.com/input/?i=plot+y%3D+2%2F3+%E2%88%9A%2818-x^2%29%2C+y+%3D+-2%2F3+%28x-3%29%2B2