Question

NO LINKS!! Please help me with this problem​

Answer 1

Answer:

$\dfrac{y^2}{36}-\dfrac{x^2}{16}=1$

Step-by-step explanation:

Standard form equation of a vertical hyperbola

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

where:

• center = (h, k)
• vertices = (h, k±a)
• co-vertices = (h±b, k)
• foci = (h, k±c) where c² = a² + b²
• $\textsf{asymptotes}: \quad y =k \pm \left(\dfrac{a}{b}\right)(x-h)$
• Transverse axis: x = h
• Conjugate axis: y = k

From inspection of the graph:

• center = (0, 0)  ⇒ h = 0, k = 0
• vertices = (0, 6) and (0, -6)  ⇒  a = 6
• co-vertices = (4, 0) and (-4, 0)  ⇒  b = 4

Substitute the found values into the formula:

$\implies \dfrac{(y-0)^2}{6^2}-\dfrac{(x-0)^2}{4^2}=1$

$\implies \dfrac{y^2}{36}-\dfrac{x^2}{16}=1$

Answer 2

${\qquad\qquad\huge\underline{{\sf Answer}}}$

The given figure shows a vertical hyperbola with its centre at origin, and as we observe the figure, we can conclude that :

Length of transverse axis is :

$\qquad \sf \dashrightarrow \: 2b = 12$

$\qquad \sf \dashrightarrow \: b = 6$

length of conjugate axis is :

$\qquad \sf \dashrightarrow \: 2a = 8$

$\qquad \sf \dashrightarrow \: a = 4$

Equation of hyperbola ~

$\qquad \sf \dashrightarrow \: \cfrac{ {y}^{2} }{ {b}^{2} } - \cfrac{ {x}^{2} }{ {a}^{2} } = 1$

plug in the values ~

$\qquad \sf \dashrightarrow \: \cfrac{ {y}^{2} }{ {6}^{2} } - \cfrac{ {x}^{2} }{ {4}^{2} } = 1$

$\qquad \sf \dashrightarrow \: \cfrac{ {y}^{2} }{ {36}^{} } - \cfrac{ {x}^{2} }{ {16}^{} } = 1$