At what point in the first quadrant on the parabola y=4-x^2 does the tangent line, together with the coordinate axis, determine a triangle of minimum area?

Answer 1

Let the point of contact of the tangent be P(a,b)
then b = 4 - a^2

we know dy/dx = -2x, which is the slope of the tangent
so at P the slope is -2a
equation of the tangent is
y = -2ax + k
but (a,b) lies on it
b = -2a(a) + k
k = b + 2a^2 = 4-a^2 + 2a^2
k = 4 + a^2

tangent equation : y = -2ax + 4+a^2

the base of the triangle is the x-intercept
the height of the triangle is the y-intercept

x-intercept: let y = 0
0 = -2ax + 4 + a^2
x = (4 + a^2)/(2a)

y-intercept: let x = 0
y = 4 + a^2

area = (1/2)(4+a^2)/(2a) (4+a^2) = (4+a^2)^2/(4a)

your turn!

Take the derivative of this using the quotient rule
Set that equal to zero and solve for a,
Sub back into the top to get b for the point P(a,b)
Sub into “area” to get the minimum area.

I got a = 2/√3