How can i find the point in the first quadrant on the parabola $$ y = 4-x ^ 2 $$ such that the triangle tangent to the parabola at the point and the coordinate axes has minimum area.
Some help to interpret the equation so minimize in this exercise i stuck i know how to calculate minima but i can't fin the equation to minimize.
I tried but $y = 4 - x^2$ and the tangent is the derivative then i have $dy/dx =-2x$ and i stuck here.Thanx for you hints and help.
On
So you calculated the derivative, which means you know the slope of the tangent line for some general $x$-value. For a given point $(x_0, y_0)$ on this parabola (so in particular, $y_0 = 4-x_0^2$), what is the equation of the tangent line to this point?
Next, given the equation of the tangent line, where does it intersect the $x$- and $y$-axes? This will give you the length of the two legs of the right triangle enclosed by the tangent line.
Finally, what is the area of such a triangle as a function of the $x$-coordinate of the point $(x_0, y_0)$?
How can you minimize this area?
ANSWER;
At the point (a,b), using the equation of the parabola
$$ b = 4-a^2$$
The equation of the tangent line is then
$$ y-b = -2a(x-a)$$
$$ y = b -2ax +2a^2$$
Substituting the value of b from the first equation,
$$ y = 4-a^2 - 2ax +2a^2 = 4 + a^2 - 2ax$$
For the point in X-axis,say $(x_1,0)$, $$ 0 = 4+a^2 - 2ax_1$$
$$ x_1 = \frac{4+a^2}{2a}$$ For the point in Y-axis, say $(0,y_1)$
$$ y_1 = 4+a^2$$
Area of the triangle $$= \frac{1}{2}x_1y_1$$
Substituting the value of $x_1,y_1$ in terms of a
$$ A = \frac{(4+a^2)^2}{4a}$$
Now set $\frac{dA}{da} = 0$
If you take the derivative using quotient rule, you get $$a = \frac{2}{\sqrt{3}}$$ and $$ b = \frac{8}{3}$$