Find an equation in standard form for the hyperbola with

Question

vertices at (0, ±2) and foci at (0, ±7).

Answer 1

Given hyperbola

General equation of a hyperbola

  We have foci and vertices are on the y-axis, which means that we needs the formula for a up and down hyperbola.

This means that the center (h, k) must be a the origin, or (0, 0). So, let's label that...

h = 0
k = 0

 We know that  "a" is the distance from your vertex and "c" is the distance from your foci

a = 2 and c = 7

We have a formula that

Now fill that

h = 0

k = 0

Therefore the required equation of hyperbola is

Answer 2

The vertices of the hyperbola are (0, 2) and (0, - 2) and its foci are (0, 7) and (0, - 7).

Since the x - coordinate is constant in the vertices and foci, this is a vertical hyperbola.

The standard form of  vertical hyperbola (y - k)²/a² - (x - h)²/b² = 1.

• Where, "b " is the number in the denominator of the positive term, If the x - term is negative, then the hyperbola is vertical.
• a = semi - transverse axis , b = semi - conjugate axis .
• Center: (h, k )
• Vertices: (h , k + a ) and (h, k - a).
• Foci: (h , k +c) and (h, k - c).

So, the x coodinate of the center of hyperbola is  0.

vertices : (0, 2) and (0,- 2)

k + a = 2 ----> (1)

k - a = - 2 ---> (2)

Add the equations (1) & (2).

2k = 0

⇒ k = 0

So,the y coordinate of center is 0.

Substitute the k value in (1),

0 + a = 2

⇒ a = 2.

foci : (0, 7) and (0, - 7)

k + c = 7

0 + c = 7

c = 7

c² = a² + b²

(7)² = (2)² + b²

49 - 4 = b²

b = √45

Substitute the (h , k), a, and b in standard form of hyperbola equation .

(y - 0)²/2² - (x - 0)²/(√45)² = 1

(y - 0)²/4 - (x - 0)²/45 = 1.

Therefore, the standard form of hyperbola is (y - 0)²/4 - (x - 0)²/45 = 1.