Determining the coordinate of C to minimize the area of a triangle ABC

Question

Given $A=(0,-10)$ and $B=(2,0)$. Determine the coordinate of $C$ in the curve $y=x^2$ which minimalize the area of triangle $ABC$.

Answer 1

The surface is half the base $AB$ times the height, where the height is the distance between the point $(x,x^2)$ and the line $AB$.

So the question is the same as asking the minimum distance of the point $(x,x^2)$ to the line $5x + y -10 = 0$.

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1

As the curve $y=x^2$ and the line $y=5x-10$ have no intersection, we can find that point where the tangent of $y=x^2$ is the same as $AB$, thus solve $$2x = 5,$$ so we get $$C \Big(\frac{5}{2},\frac{25}{4} \Big)$$

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2

The distance of a point $(x,y)$ and that line is given by $$\frac{|5x+x^2-10|}{\sqrt{125}},$$ which can be written as $$\frac{\left|\Big(x - \tfrac{5}{2} \Big)^2 + \tfrac{15}{4}\right|}{\sqrt{125}},$$ so the minimum is for $x = \tfrac{5}{2}$ and we get $$C \Big(\frac{5}{2},\frac{25}{4} \Big)$$

Answer 2

HINT:

Any point on the curve $y=x^2$ can be expressed as $(t,t^2)$

So, $\triangle ABC$ will be the absolute value of $$\frac12\begin{vmatrix}0 & 2 & t\\-10 &0 &t^2\\1 &1&1\end{vmatrix}$$

On simplification this will be a Quadratic equation in $t$

Now, for $\displaystyle A>0, Ax^2+Bx+C=\frac{(2Ax+B)^2+4CA-B^2}{4A}\ge\frac{4CA-B^2}{4A}$