Question

The equation of the ellipse whose vertices are (-4, 3), (8, 3) and whose eccentricity is 5/6 is

• A. $\frac{(x + 2)^{2}}{36} + \frac{(y + 3)^{2}}{11} = 1$
• B. $\frac{(x - 2)^{2}}{36} + \frac{(y + 3)^{2}}{11} = 1$
• C. $\frac{x^{2}}{36} + \frac{y^{2}}{11} = 1$
• D. $\frac{(x - 2)^{2}}{36} + \frac{(y - 3)^{2}}{11} = 1$

Answer 1

Answer: (D)

$\frac{(x - 2)^{2}}{36} + \frac{(y - 3)^{2}}{11} = 1$

Answer 2

Given A = (8, 3) A’ = (-4, 3)

∴ Major axis parallel to x-axis.

AA’ = 12 = 2a ⇒ a = 6

Centre = (α, β) = mid point of AA’ = (2, 3)

Given e =5/6.

⇒ $\frac{a^{2} - b^{2}}{a^{2}} = \frac{25}{36}$ ⇒ b = $\sqrt{11}$

∴ Required equation of ellipse is $\frac{(x - \alpha)^{2}}{a^{2}} + \frac{(y - \beta)^{2}}{b^{2}} = 1$ ie., $\frac{(x - 2)^{2}}{36} + \frac{(y - 3)^{2}}{11} = 1$