Answer:
The standard form of a hyperbola with vertices and foci on the x-axis:
[tex]\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1[/tex]
where:
- center: (h, k)
- vertices: (h+a, k) and (h-a,k)
- Foci: (h+c, k) and (h-c, k) where the value of c is c² = a² + b²
- Slopes of asymptotes: [tex]\pm\left(\dfrac{b}{a}\right)[/tex]
Part 1
The center of the given hyperbola is (0, 0), therefore:
[tex]\implies \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1[/tex]
Therefore [tex](\pm a,0)[/tex] are the vertices. From inspection of the graph, [tex]a=2[/tex].
Part 2
Choose two points on the asymptote with the positive slope:
(0, 0) and (4, 6)
Use the slope formula to find the slope:
[tex]\sf slope\:(m)=\dfrac{change\:in\:y}{change\:in\:x}=\dfrac{6-0}{4-0}=\dfrac{3}{2}[/tex]
Part 3
Use the slopes of asymptotes formula, compare with the slope found in part 2:
[tex]\implies \dfrac{b}{a}=\dfrac{3}{2}[/tex]
Therefore, [tex]b=3[/tex]
Part 4
Substitute the found values of [tex]a[/tex] and [tex]b[/tex] into the equation from part 1:
[tex]\implies \dfrac{x^2}{2^2}-\dfrac{y^2}{3^2}=1[/tex]
[tex]\implies \dfrac{x^2}{4}-\dfrac{y^2}{9}=1[/tex]
