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Pachacha [2.7K]
1 year ago
10

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

Mathematics
1 answer:
SpyIntel [72]1 year ago
4 0

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)

<h3><u>Part 1</u></h3>

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.

<h3><u>Part 2</u></h3>

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}

<h3><u>Part 3</u></h3>

Use the <u>slopes of asymptotes</u> formula, compare with the slope found in part 2:

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

Therefore, b=3

<h3><u>Part 4</u></h3>

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

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