Question

7 an ellipse has a vertex at (0, -7), a co - vertex at (4, 0), and a center at the origin. which is the equation of the ellipse in standard form?
options:
\\(\frac{x^2}{49}+\frac{y^2}{16}=1\\)
\\(\frac{x^2}{4}+\frac{y^2}{7}=1\\)
\\(\frac{x^2}{7}+\frac{y^2}{4}=1\\)
\\(\frac{x^2}{16}+\frac{y^2}{49}=1\\)

Explanation:

Step1: Identify ellipse orientation

The vertex $(0, -7)$ lies on the y-axis, so the major axis is vertical. The standard form for this ellipse (centered at the origin) is $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$, where $a$ is the distance from the center to a vertex, and $b$ is the distance from the center to a co-vertex.

Step2: Find $a$ value

The vertex is at $(0, -7)$, so $a = |-7 - 0| = 7$. Then $a^2 = 7^2 = 49$.

Step3: Find $b$ value

The co-vertex is at $(4, 0)$, so $b = |4 - 0| = 4$. Then $b^2 = 4^2 = 16$.

Step4: Substitute into standard form

Substitute $a^2=49$ and $b^2=16$ into the vertical major axis ellipse equation: $\frac{x^2}{16}+\frac{y^2}{49}=1$.

Answer:

$\boldsymbol{\frac{x^2}{16}+\frac{y^2}{49}=1}$