Question

Which is the equation of an ellipse centered at the origin with foci on x-axis, major axis of length 12 and minor axis of length 8?

Answer 1

Answer:

$\frac{x^{2} }{36}+\frac{y^2}{16}=1$

Step-by-step explanation:

Center of the ellipse is at origin (0, 0)

So (h, k) = (0, 0)

Length of major axis = 12

2a = 12

⇒ a = 6

Length of minor axis = 8

⇒ 2b = 8

⇒ b = 4

Since focus is on the x-axis, so it's a horizontal ellipse.

Equation of the horizontal ellipse is in the form of,

$\frac{(x-h)^{2} }{a^2}+\frac{(y-k)^2}{b^2}=1$

By substituting the given values of a, b, and (h, k)

$\frac{(x-0)^{2}}{6^2}+\frac{(y-0)^2}{4^2}=1$

$\frac{x^{2} }{36}+\frac{y^2}{16}=1$