Question

An ellipse has vertices at (0, #17) and foci at (0, ±15). Write the equation of the ellipse in standard form. Graph the ellipse.

Answer 1

ANSWER

$\frac{ {x}^{2} }{ 64 } + \frac{ {y}^{2} }{ 289 } = 1$

See attachment for the graph

EXPLANATION

The standard equation of the vertical ellipse with center at the origin is given by

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

where

${a}^{2} \: > \: {b}^{2}$

The ellipse has its vertices at (0,±17).

This implies that:a=±17 or a²=289

The foci are located at (0,±15).

This implies that:c=±15 or c²=225

We use the following relation to find the value of b²

${a}^{2} - {b}^{2} = {c}^{2}$

$\implies \: 289 - {b}^{2} = 225$

$- {b}^{2} = 225 - 289$

$- {b}^{2} = - 64$

${b}^{2} = 64$

We substitute into the formula for the standard equation to get:

$\frac{ {x}^{2} }{ 64 } + \frac{ {y}^{2} }{ 289 } = 1$