### Instruction: Points $A$ and $B$ are on parabola $y=4x^2+7x-1$, and the origin is the midpoint of $\overline{AB}$. Find the square of the length of $\overline{AB}$.### Response: I notice that the problem is asking for the square of the length of $\overline{AB}$, not the actual length. This might make things easier, since I can avoid taking square roots.
I also notice that the origin is the midpoint of $\overline{AB}$, which means that the coordinates of $A$ and $B$ must have opposite signs and the same magnitude. For example, if $A$ is $(3,10)$, then $B$ must be $(-3,-10)$.
So, I can write the coordinates of $A$ as $(a,4a^2+7a-1)$ and the coordinates of $B$ as $(-a,4a^2-7a-1)$.
To find the square of the length of $\overline{AB}$, I need to use the distance formula, which says that the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.
Squaring this formula, I get that the square of the distance is $(x_2-x_1)^2+(y_2-y_1)^2$.
Plugging in the coordinates of $A$ and $B$, I get that the square of the length of $\overline{AB}$ is $((-a)-a)^2+(4a^2-7a-1-(4a^2+7a-1))^2$.
Simplifying this expression, I get $4a^2+(-14a)^2=4a^2+196a^2=200a^2$.
So, the square of the length of $\overline{AB}$ depends on the value of $a$, which I have not determined yet.
To find $a$, I need to use the fact that the origin is the midpoint of $\overline{AB}$.
This means that the average of the x-coordinates of $A$ and $B$ is zero, and the average of the y-coordinates of $A$ and $B$ is zero.
So, I have the equations $\frac{a+(-a)}{2}=0$ and $\frac{4a^2+7a-1+(4a^2-7a-1)}{2}=0$.
The first equation is always true, so it does not help me.
The second equation simplifies to $4a^2-1=0$, which I can solve by adding $1$ to both sides and dividing by $4$.
I get $a^2=\frac{1}{4}$, which has two solutions: $a=\frac{1}{2}$ or $a=-\frac{1}{2}$.
Either solution will work, since the parabola is symmetric about the y-axis.
So, I can plug in $a=\frac{1}{2}$ into the expression for the square of the length of $\overline{AB}$, which is $200a^2$.
I get $200(\frac{1}{2})^2=200\cdot \frac{1}{4}=50$.
Therefore, the square of the length of $\overline{AB}$ is 50.