Eliminate $x_1,x_2,y_1,y_2,c$ from the following equations

Question

I need to eliminate $x_1,x_2,y_1,y_2,c$ from the following equations.What would be the correct ( and quick ) technique to do so?

$x_1y_1=1$

$y_1=4x_1+c$

$x_2y_2=1$

$y_2=4x_2+c$

$h=(2x_1+x_2)/3$

$k=(2y_1+y_2)/3$

I could get $h=(2x_1+x_2)/3$ and $k=((2x_2+x_1)/3x_1x_2)$

After that what to do?

Answer 1

$$y_1- 4x_1 = y_2 - 4x_2$$

which gives

$$\frac{1}{x_1} - 4x_1 = \frac{1}{x_2} - 4x_2$$

which gives

$$-1/x_1x_2 = 4 \quad \quad \quad (1)$$

(leaving aside the case when $x_1 = x_2$ )

Put this value of $x_1x_2$ in the equation for $k$ you got, and then solve those two equations to get the value of $x_1$ and $x_2$ in terms of $h$ and $k$ and then put these in equation $(1)$ to get

$$(k + 8h)(2h+k) = 2$$

Answer 2

HINT.- $(x_1,y_1)$ and $(x_2,y_2)$ are the two points of intersection of the line $y=4x+c$ and the hyperbola $yx=1$ One has $$x_1,x_2=\frac{-c\pm\sqrt{c^2+16}}{8}\\y_1,y_2=\frac{c\pm\sqrt{c^2+16}}{2}$$ You have to eliminate $c$ from the system $$24h=-3c\pm\sqrt{c^2+16}\\6k=3c\pm\sqrt{c^2+16}$$ I get $$128h^2+80hk+8k^2-16=0$$ which apparently differs from Akash's result above and then one of the two would have made an error in the calculation.I bet I who he was wrong.

NOTE.- I was who have made the (quite distracted) mistake, puting $-7k^2$ instead of $8k^2$ which is the result of $9k^2-k^2$. Now both answers are correct.