Question

For the ellipse $\frac{x^2}{4}+\frac{y^2}{25}=1$, the vertices are the points _____ and _____.

Short answer

For the ellipse $\frac{x^2}{4}+\frac{y^2}{25}=1$, the vertices are the points $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{-}\mathbf{5}\mathbf{)}and\mathbf{}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{5}\mathbf{)}$.

Step-by-step solution

Step 1. Given Information.

We want to fill the gap in the given equation with the missing terms.

Step 2. Method to find:

We know that

If an ellipse is given with the following equation:

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

where $a>b$, then it is centred at the origin, the major axis is the y-axis and the parameters $aandb$represent its width $2a$and its height $2b$.

Since the vertices lie on the major axis and the segment connecting them is the width of the ellipse, then it is $2a$long.

Therefore, the vertices are the points$(0,-a)and(0,a)$

Step 3. Solution:

The following equation is given

$\frac{x^2}{4}+\frac{y^2}{25}=1$

then by comparing it to the equation of an ellipse given in Step 2. we can see that

Since $2^2=4$and because $25>4\iff 5>2$

Therefore, the vertices are points:

$V_1=(0,-5)and{V}_2=(0,5)$