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# Find an equation in standard form for the hyperbola with vertices at (0, ±2) and foci at (0, ±7)?

**Solution:**

Given:

**Vertices at (0, ±2) and foci at (0, ±7)**

As both vertices and foci lie on the y-axis

Equation of hyperbola is y^{2}/b^{2} - x^{2}/a^{2} = 1

**Vertices = (0, ±2)**

b = 2

**Foci = (0, ±7)**

be = 7

Eccentricity e = 7/2

**We know that,**

e = √(1 + a^{2}/b^{2})

**Substituting the values**

7/2 = √(1 + a^{2}/b^{2})

**Squaring on both sides**

49/4 = 1 + a^{2}/b^{2}

⇒ a^{2}/b^{2} = 49/4 - 1

⇒ a^{2/}b^{2} = 45/4

Here,

a^{2} = 45/4 × 2^{2}

⇒ a^{2} = 45/4 × 4

⇒ a^{2} = 45

⇒ a = 3√5

**So the equation of hyperbola is**

y^{2}/4 - x^{2}/45 = 1

**Therefore, the equation of hyperbola in standard form is y ^{2}/4 - x^{2}/45 = 1.**

## Find an equation in standard form for the hyperbola with vertices at (0, ±2) and foci at (0, ±7)?

**Summary:**

An equation in standard form for the hyperbola with vertices at (0, ±2) and foci at (0, ±7) is y^{2}/4 - x^{2}/45 = 1.

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