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Points $A$ and $B$ are on parabola $y=3x^2-5x-3$, and the origin is the midpoint of $\overline{AB}$. Find the square of the length of $\overline{AB}$.

 Jul 21, 2019
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Points\( A\) and \(B\) are on parabola \(y=3x^2-5x-3\), and the origin is the midpoint of \(\overline{AB}\).
Find the square of the length of \(\overline{AB}\).

 

\(\text{ Let $A(x_A,\ y_A)$ and $B(x_B,\ y_B)$ }\)

 

\(\begin{array}{|lrcll|} \hline (1) & y_A&=&3x_A^2-5x_A-3 \\ (2) & y_B&=&3x_B^2-5x_B-3 \\ (3) & x_B&=&-x_A\\ (4) & y_B&=&-y_A\\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline (1) & y_A&=&3x_A^2-5x_A-3 \\ \hline (2) & y_B&=&3x_B^2-5x_B-3 \quad &| \quad y_B=-y_A,\ x_B=-x_A \\ & -y_A&=&3(-x_A)^2-5(-x_A)-3 \\ (2)^* & -y_A&=&3x_A^2+5x_A-3 \\ \hline (1)+(2)^*: & y_A-y_A&=&3x_A^2-5x_A-3+3x_A^2+5x_A-3\\ & 0&=&6x_A^2-6 \quad &| \quad : 6 \\ & 0&=&x_A^2-1 \\ & x_A^2 &=& 1 \\ & \mathbf{x_A} &=& \mathbf{1} \\ & \mathbf{x_B} &=& -x_A = \mathbf{-1} \\ \hline & y_A &=& 3x_A^2-5x_A-3 \\ & &=& 3(1)^2-5(1)-3 \\ & &=& 3-5-3 \\ & \mathbf{y_A} &=& \mathbf{-5} \\ & \mathbf{y_B} &=& -y_A = -(-5) = \mathbf{5} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline A(1,\ -5),\ B(-1,\ 5) \\\\ \text{The square of the length of } \overline{AB} : \\ \left(1-(-1)\right)^2 + \left(-5-(5)\right)^2\\ =2^2+(-10)^2 \\ =4+100 \\ \mathbf{= 104} \\ \hline \end{array} \)

 

laugh

 Jul 22, 2019
edited by heureka  Jul 22, 2019

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