Minimizing area of a triangle with two fixed point and a point on parabola

Question

A triangle is made up of three points, $A, B$, and $P$.

$A(-1, 0)$

$B(0, 1)$

$P$ is a point on $y^2 = x$

Minimize the area of Triangle $ABP$.

My approach is far too complicated, which generalize $P$ as $(t^2, t)$, which makes the whole thing quite messy. Maybe that's the right thing to do but I'm looking at the whole thing wrong.

Answer 1

HINT:

$1$st of all, it should be $(t^2,t)$

Now the area $$\frac12\left|\det\begin{pmatrix} -1 & 0 & 1 \\ 0 & 1 & 1 \\ t^2 & t&1\end{pmatrix}\right|$$

$$=\frac12 (t^2-t+1)=\frac{(2t-1)^2+3}8\ge \frac38$$

So, the area will be minimum if $t=\frac12$

Answer 2

One way is with lagrange multipliers. Think of $ AB $ as the base of the triangle. Then the area will be minimized when P is closest to the line $ AB $. This requires the tangent of $ y^2 = x $ to be parallel to the line $ AB $, which has slope 1. So $ \frac{dy}{dx} = \frac{1}{2 \sqrt{x}} = 1 $, which gives $ x = \frac{1}{4} $.