Question

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

Answer 1

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

Explanation:

This is an Ellipse with Vertical Major Axis

with the equation form `(x-h)^2/b^2+(y-k)^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(h, k+a)=V_1(5, -1+4)=V_1(5, 3)`
Vertex `V_2(h, k-a)=V_2(5, -1-4)=V_2(5, -5)`

Co-vertex `V_3(h+b, k)=V_3(5+3,-1)=V_3(8, -1)`
Co-vertex `V_4(h-b, k)=V_3(5-3, -1)=V_3(2, -1)`

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

Kindly see the graph of `(x-5)^2/9+(y+1)^2/16=1` and locate the vertices and co-vertices

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