ABC is a triangle, right-angled at B. And it is inscribed in the parabola $y^2=4x$. Find the minimum length of AC. We have a right-angled triangle inscribed in the parabola $y^2=4x$ and we have to find the minimum length of its hypotenuse. 
Taking the points as $A((t_1)^2,2t_1)$, $B((t_2)^2,2t_2)$ and $C((t_3)^2,2t_3)$. 
We know that $\overrightarrow{AB}$ is perpendicular to $\overrightarrow{BC}$. Hence the product of their slopes is $-1$. 
So, we get 
$$(t_1+t_2)(t_2+t_3)=-4$$
But I'm not sure how to proceed. 
 A: Claim: The required shortest length is AB = 8 units where A is the intersection of $L_1:y = x$ and $C: y^2 = 4x$ and B is similarly defined using $L_2: y = -x$.

Clearly, $\angle AOB = 90^0$.
The case when both endpoints of its hypotenuse lying further right of x = 4 will never be qualified.
Any other possible candidates with one endpoint of its hypotenuse lying further right of x = 4 will also be not qualified. In the illustration, $\triangle A’b’O’$ is an example with A’b’ > AB. (I am not going to spend time in proving this part.)
Now, let us test if a (governed) right angled triangle can be form when both A’ and B’ are on the left side of x = 4. That is, $A’ = (p, 2\sqrt p)$ and B’ = $(q, -2 \sqrt q)$ where $0 \lt p, q \lt 4$.
Using A’B’ as diameter, form the dotted circle. By the converse of “angles in the semi-circle”, the third point (let us call it X) of the right-angled triangle A’XB’ must lie outside the parabola. This means no other right angled triangles meeting the requirement can be formed. This can also be proved by:-


*

*Equation of the circle XA’B’ is $D: (x – p)(x – q) + (y – 2\sqrt p)(y + 2 \sqrt q) = 0$ 

*The “tangent length” from the extreme point (0, 0) to D is imaginary.
