An object enters the solar solar system. Also the sun is at the origin of the path.
The equation of the entering object is given as,
[tex]\begin{gathered} y\text{ = }2x\text{ -2 } \\ y\text{ = 2 ( x - 1)} \end{gathered}[/tex]
The equation for departing the solar system is given as,
[tex]\begin{gathered} y\text{ = -2x + 2} \\ y\text{ = -2 ( x - 1 )} \end{gathered}[/tex]
The combined equation for the object is given as,
[tex]y\text{ = }\pm2(x-1)[/tex]
The equation of the asymptote is given as,
[tex]y\text{ = }\pm\frac{b}{a}(\text{ x -h )+k}[/tex]
Comparing the given equation with the asymptote we get,
[tex]\begin{gathered} \frac{b}{a}\text{ = 2} \\ h\text{ = 1} \\ k\text{ = 0} \end{gathered}[/tex]
The distance from the sun is 0.5 au .
Therefore,
[tex]\begin{gathered} a\text{ = 0.5} \\ \frac{b}{0.5}\text{ = 2} \\ b\text{ = 2 }\times\text{ 0.5} \\ b\text{ = 1} \end{gathered}[/tex]
The general equation for the hyperbola symmetric to x axis is given as,
[tex]\frac{(x-h)^2}{a^2}\text{ + }\frac{(y-k)^2}{b^2}\text{ = 1}[/tex]
Substituting the given values in the equation ,
a = 0.5 , b = 1 , h = 1 and k = 0.
[tex]\begin{gathered} \frac{(x-1)^2}{0.5^2}+\frac{(y-0)^2}{1^2}\text{ = 1} \\ \frac{(x-1)^2}{0.25}+\frac{y^2}{1^2}\text{ = 1} \\ 4(x-1)^2+y^2\text{ = 1} \end{gathered}[/tex]
Thus the required equation is ,
[tex]4(x-1)^2+y^2\text{ = 1}[/tex]
The graph is given as:
