Question

NO LINKS!! Please help me with this problem​

Answer 1

`{\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`

Answer 2

`(y^2)/(36)-(x^2)/(16)=1`

Explanation:

Standard form equation of a vertical hyperbola

`((y-k)^2)/(a^2)-((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((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 ((y-0)^2)/(6^2)-((x-0)^2)/(4^2)=1`

`\implies (y^2)/(36)-(x^2)/(16)=1`