If a hyperbola is centered at the origin and has vertex and focus on the x-axis, then its equation is
[tex]\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1.[/tex]
1. The vertex A of hyperbola has coordinates (-6,0), then the distance from the origin (center of the hyperbola) to the vertex is
[tex]a=6.[/tex]
2. The focus F has coordinates (10,0), then the distance from the origin to the focus is
[tex]c=10.[/tex]
3. Find b, using formula [tex]c^2=a^2+b^2:[/tex]
[tex]10^2=6^2+b^2,\\ \\b^2=100-36=64,\\ \\b=8.[/tex]
4. The equation of the hyperbola is
[tex]\dfrac{x^2}{36}-\dfrac{y^2}{64}=1.[/tex]
5. The directrices' equations are
[tex]x=-\dfrac{a}{e},\ x=\dfrac{a}{e},[/tex]
where [tex]e=\dfrac{c}{a}.[/tex]
In your case,
[tex]e=\dfrac{10}{6}[/tex]
and directrices' equations are
[tex]x=-\dfrac{6}{\frac{10}{6}}=-3.6,\ x=\dfrac{6}{\frac{10}{6}}=3.6.[/tex]
Answer: x=-3.6, x=3.6, correct choice is C
