Question

Al what lirst-quadrant point on the parabola y =4- X does the tangent, together with the coordinate
axes, determine a triangle of minimum area.
Ans. (2V3/3, 8/3)

Answer 1

at the point (a,b) on the parabola, the tangent has slope -2a, so the tangent line has equation
y = -2a(x-a) + b = -2a(x-a) + 4-a^2 = -2ax + a^2+4
Now just find the x- and y-intercepts of that line. The area is
A = (a^2+4)^2 / (4a)
dA/da = (3a^4+8a^2-16)/(4a^2)
minimum area occurs where dA/da = 0, at a^2 = (-8±16)/6 = -4 or 4/3
Thus, since we want a in the 1st quadrant, that gives a = 2/√3
and b = 4-a^2 = 8/3

Answer 2

You don't have a parabola, I will assume your question
might sound like this:

<b>At what first-quadrant point on the parabola y =4 - x^2 does the tangent,
together with the coordinate axes, determine a triangle of minimum area.</b>

Now it becomes a doable question, but your title of Pre-Calculus will make it hard to do without Calculus.

Make a sketch. Let A(a,b) be the point we are looking for.
Since A lies on the parabola, we can label it A(a, 4 - a^2)

Here is where we need Calculus.
dy/dx = -2x
so at A, the slope of the tangent will be -2a

equation of tangent:
y - (4-a^2) = -2a(x-a)
y - 4 + a^2 = -2ax + 2a^2
2ax + y = a^2 + 4

This is the equation of the tangent at A

find the x and y intercepts, so we get the base and height of our triangle.
let x = 0, then y = a^2 + 4
let y = 0 , then x = (a^2 + 4)/(2a)

Area of triangle : A = (1/2)(base)(height)
A = (1/2)(a^2 + 4)(a^2 + 4) / (2a)
= (a^4 + 8a^2 + 16)/(4a)
= (1/4)a^3 + 2a + 4a^-1

dA/da = (3/4)a^2 + 2 - 4/a^2 = 0 for a min of A
times a^2
(3/4)a^4 + 2a^2 - 4 = 0
times 4
3a^4 + 8a^2 - 16 = 0
a = 2/√3 = 2√3/3 , since a > 0 in quadrant I
the y value would be 4 - (2/√3)^2 = 8/3

I agree with your answer.

(that was fun)