Question

1.- What is 'a' for this hyperbola?

Answer 1

Answer:

The standard form of a hyperbola with vertices and foci on the x-axis:

$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$

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: $\pm\left(\dfrac{b}{a}\right)$

Part 1

The center of the given hyperbola is (0, 0), therefore:

$\implies \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$

Therefore $(\pm a,0)$ are the vertices.  From inspection of the graph, $a=2$.

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:

$\sf slope\:(m)=\dfrac{change\:in\:y}{change\:in\:x}=\dfrac{6-0}{4-0}=\dfrac{3}{2}$

Part 3

Use the slopes of asymptotes formula, compare with the slope found in part 2:

$\implies \dfrac{b}{a}=\dfrac{3}{2}$

Therefore, $b=3$

Part 4

Substitute the found values of $a$ and $b$ into the equation from part 1:

$\implies \dfrac{x^2}{2^2}-\dfrac{y^2}{3^2}=1$

$\implies \dfrac{x^2}{4}-\dfrac{y^2}{9}=1$