# How do I graph the ellipse with the equation (x−5)^2/9+(y+1)^2/16=1?

From the equation, by inspection, Center$\left(5 , - 1\right)$ length of Semi-major axis$= 4$, length of Semi-minor axis$= 3$, Vertices at $\left(5 , 3\right)$ and $\left(5 , - 5\right)$, Co-vertices at $\left(8 , - 1\right)$ and $\left(2 , - 1\right)$

#### Explanation:

This is an Ellipse with Vertical Major Axis

with the equation form ${\left(x - h\right)}^{2} / {b}^{2} + {\left(y - k\right)}^{2} / {a}^{2} = 1$

so that ${a}^{2} = 16$ and $a = 4$
and ${b}^{2} = 9$ and $b = 3$

so we can calculate, Vertices using the values of $a , b , h , k$

Vertex ${V}_{1} \left(h , k + a\right) = {V}_{1} \left(5 , - 1 + 4\right) = {V}_{1} \left(5 , 3\right)$
Vertex ${V}_{2} \left(h , k - a\right) = {V}_{2} \left(5 , - 1 - 4\right) = {V}_{2} \left(5 , - 5\right)$

Co-vertex ${V}_{3} \left(h + b , k\right) = {V}_{3} \left(5 + 3 , - 1\right) = {V}_{3} \left(8 , - 1\right)$
Co-vertex ${V}_{4} \left(h - b , k\right) = {V}_{3} \left(5 - 3 , - 1\right) = {V}_{3} \left(2 , - 1\right)$

We don't use the Foci to graph the ellipse so there's no need to compute for them

Kindly see the graph of ${\left(x - 5\right)}^{2} / 9 + {\left(y + 1\right)}^{2} / 16 = 1$ and locate the vertices and co-vertices

graph{(x-5)^2/9+(y+1)^2/16=1[-12,12,-6,6]}