Question

The ellipse with x-intercepts (2, 0) and (-2, 0); y-intercepts (0, 4) and (0, -4).

Answer 1

Answer:

$\dfrac{x^2}{4}+\dfrac{y^2}{16}=1$

Step-by-step explanation:

If the ellipse has its x-intercepts at points (2, 0) and (-2, 0) and y-intercepts at points (0, 4) and (0, -4), then its symmetric across the y-axis and across the x-axis.

Moreover,

$a=2\\ \\b=4$

The equation of such ellipse is

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

Hence, the equation of the ellipse is

$\dfrac{x^2}{2^2}+\dfrac{y^2}{4^2}=1\\ \\ \\\dfrac{x^2}{4}+\dfrac{y^2}{16}=1$