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

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 $ABC$.

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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 $5x + y -10 = 0$.

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 $$C \Big(\frac{5}{2},\frac{25}{4} \Big)$$

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 $$C \Big(\frac{5}{2},\frac{25}{4} \Big)$$

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Thank you very much for the answer. It does help. You have just shown me other ways to answer it. – Monica Sendi Afa Jul 5 '14 at 10:11
@Monica: Your welcome! – johannesvalks Jul 5 '14 at 10:15

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

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Does it mean that the area of triangle $ABC$ is $A = 1/2 * (2*t^2-9*t+20)$ ? In order to find the minimum area, $A'=0$ right? – Monica Sendi Afa Jul 5 '14 at 9:33
@MonicaSendiAfa, I found $t^2-5t+10$ which $=\dfrac{(2t-5)^2+40-15}4\ge\dfrac{40-15}4$ So, the minimum value occurs if $2t-5=0$. We can use mathworld.wolfram.com/SecondDerivativeTest.html as well – lab bhattacharjee Jul 5 '14 at 9:37
Thank you for the answer and explanation. My apologize for my mistakes. Yes, of course the area is $t^2-5t+10$. However, based on your explanation in answer "for $A>0, Ax^2+Bx+C=[(2Ax+B)^2+4CA=B^2]/(4A)$", shouldn't $t^2-5t+10 = [(2t-5)^2+40-25]/4$ ? Well, actually it doesn't really matter since to find the answer, we only need to find the value of $t$ satisfying $2t-5=0$, right? – Monica Sendi Afa Jul 5 '14 at 9:53
@MonicaSendiAfa, Right. Sorry for the typo – lab bhattacharjee Jul 5 '14 at 9:55
It's alright. I do thank you for the help. Anyway, I recall my high school teacher once explained that to find the minimum or maximum value of a function, the first derivative of the function is equal to zero. In this case, $dA/dt=0$ which also gives $2t-5=0$. This way could be used as well, could it? Once again, thank you for the answer. – Monica Sendi Afa Jul 5 '14 at 10:02