$8x-y=9$ intersects hyperbola, $\frac{x^2}{16}-\frac{y^2}{2023}=9$ in $A$ and $B$.The ⊥ of $AB$ intersects the axes at $C$ and $D$.Find area of $COD$.

Question

Problem

In the Cartesian rectangular system $xOy$, the line $8x - y = 9$ intersects the hyperbola

$\frac{x^2}{16} - \frac{y^2}{2023} = 9$

in $A$ and $B$. The perpendicular bisector of $AB$ intersects the axes $Ox$ and $Oy$ in $C$ and $D$. Find the area of triangle $COD$.

• This problem is from the team round of a local high school math competition that has ended

My Work

We need to find the intersection points (A and B), so we have use the two equations to solve for these coordinates.

(1) $y = 8x-9$

Plug this into the second equation.

$\frac{x^2}{16} - \frac{(8x-9)^2}{2023} = 9$

$2023x^2 - 16(64x^2 - 144x + 81) = 16 \cdot 2023 \cdot 9$

$999 x^2 + 2304 x - 292608 = 0$

This equation seems super messy, am I going on the correct path? Am I required to use the quadratic formula? After I solve for x and y, I will find the midpoint and slope to find the equation of the perpendicular bisector and then use that to calculate the area of the triangle, but I am having trouble solving for x and y currently. I appreciate any help! Thank you.

Answer 1

Hint

Let $A=(x_{1},y_{1}),B=(x_{2},y_{2})$, the midpoint of AB be $M=(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$.

We have $\frac{{x_{1}}^2}{16}-\frac{{y_{1}}^2}{2023}=9,\frac{{x_{2}}^2}{16}-\frac{{y_{2}}^2}{2023}=9$

So $(\frac{{x_{1}}^2-{x_{2}}^2}{16})-(\frac{{y_{1}}^2-{y_{2}}^2}{2023})=0$

So $\frac{y_{1}+y_{2}}{x_{1}+x_{2}}⋅\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{2023}{16}$

It is easy to see that $\frac{y_{1}-y_{2}}{x_{1}-x_{2}}$ is the slope of AB, and $\frac{y_{1}+y_{2}}{x_{1}+x_{2}}=\frac{\frac{y_{1}+y_{2}}{2}-0}{\frac{x_{1}+x_{2}}{2}-0}$, so $\frac{y_{1}+y_{2}}{x_{1}+x_{2}}$ is the slope of OM.(O=(0,0))

So $OM:y=\frac{2023}{128}x$.

Now can you know how to solve this problem?