Answer :

**Given:** Foci are (6, 4) and (-4, 4) and eccentricity is 2

**To find:** equation of the hyperbola

**Formula used:**

The standard form of the equation of the hyperbola is,

Center is the mid-point of two foci.

Distance between the foci is 2ae and b^{2} = a^{2}(e^{2} – 1)

**The distance** **between two points (m, n) and (a, b) is given by**

**Mid-point theorem:**

Mid-point of two points (m, n) and (a, b) is given by

Center of hyperbola having foci (6, 4) and (-4, 4) is given by

= (1, 4)

The distance between the foci is 2ae, and Foci are (6, 4) and (-4, 4)

{∵ e = 2}

b^{2} = a^{2}(e^{2} – 1)

The equation of hyperbola:

⇒ 12(x^{2} + 1 – 2x) – 4(y^{2} + 16 – 8y) = 75

⇒ 12x^{2} + 12 – 24x – 4y^{2} – 64 + 32y – 75 = 0

⇒ 12x^{2} – 4y^{2} – 24x + 32y – 127 = 0

Hence, required equation of hyperbola is **12x ^{2} – 4y^{2} – 24x + 32y – 127 = 0**

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