Find the equation of the hyperbola whose foci are (8,4) and (0,4) and

1.	Find the equation of the hyperbola whose foci are (8,4) and (0,4) and has am eccentricity of 2
Answer» Focus of hyperbola are (8,4) and (0,4). Hence, Focus of hyperbola lie on line y = 4, (Parallel to x-axis) ∴ Centre of the hyperbola is mid-point of both focus = (\(\frac{8+0}{2}\),\(\frac{4+4}{2}\)) = (4,4) ∵ Eccentrictiy of hyperbola is e = 2. We know that, Focus one at a distance of ae from the centre. ∵ Distance of focus (0,4) and centre (4,4) = ae ⇒ ae = 4 ⇒ a = \(\frac{4}{e}\) = \(\frac{4}{2}\) = 2. We also know that in hyperbola, e = \(\sqrt{1+\frac{b^2}{a^2}}\) ⇒ \(1+\frac{b^2}{a^2}\) = e² ⇒ b² = (e² - 1)a² ⇒ b² = (2² - 1) x 2² = (4 - 1) x 4 = 12 ⇒ b = \(\sqrt{12}\). Since, Centre of the hyperbola is (4,4) & a² = 4,b² = 12 and focus of hyperbola lies on a line parallel to x - axis. ∴ Equation of hyperbola is \(\frac{(x-4)^2}{a^2}\) - \(\frac{(y-4)^2}{b^2}\) = 1 ⇒ \(\frac{(x-4)^2}{4}\) - \(\frac{(y-4)^2}{12}\) = 1

Find the equation of the hyperbola whose foci are (8,4) and (0,4) and has am eccentricity of 2

Answer»

Focus of hyperbola are (8,4) and (0,4).

Hence,

Focus of hyperbola lie on line y = 4,

(Parallel to x-axis)

∴ Centre of the hyperbola is mid-point of both focus = (\(\frac{8+0}{2}\),\(\frac{4+4}{2}\))

= (4,4)

∵ Eccentrictiy of hyperbola is e = 2.

We know that,

Focus one at a distance of ae from the centre.

∵ Distance of focus (0,4) and centre (4,4) = ae

⇒ ae = 4

⇒ a = \(\frac{4}{e}\)

= \(\frac{4}{2}\) = 2.

We also know that in hyperbola,

e = \(\sqrt{1+\frac{b^2}{a^2}}\)

⇒ \(1+\frac{b^2}{a^2}\) = e²

⇒ b² = (e² - 1)a²

⇒ b² = (2² - 1) x 2² = (4 - 1) x 4 = 12

⇒ b = \(\sqrt{12}\).

Since,

Centre of the hyperbola is (4,4) & a² = 4,b² = 12 and focus of hyperbola lies on a line parallel to x - axis.

∴ Equation of hyperbola is \(\frac{(x-4)^2}{a^2}\) - \(\frac{(y-4)^2}{b^2}\) = 1

⇒ \(\frac{(x-4)^2}{4}\) - \(\frac{(y-4)^2}{12}\) = 1