Question

How do you find the critical points for `(9x^2)/25 + (4y^2)/25 = 1`?

Answer 1

Critical points are `(0,5/2)`, `(-5/3,0)`, `(0,-5/2)`, `(5/3,0)`,

Explanation:

A critical point of a differentiable function is any point on the curve in its domain, where its derivative is `0` or undefined.

For `(9x^2)/25+(4y^2)/25=1`, `(dy)/(dx)` is given by

`9/25xx2x+4/25xx2yxx(dy)/(dx)=0`

i.e. `(dy)/(dx)=-(18x)/25-:(8y)/25=-(9x)/(4y)`,

which is `0`, when `x=0` i.e. at `(4y^2)/25=1` or `y=+-5/2`

Further, `(dy)/(dx)` is undefined at `-(4y)/(9x)=0` or `y=0`, where `x=+-5/3`.
Hence, critical points are `(0,5/2)`, `(-5/3,0)`, `(0,-5/2)` and `(5/3,0)`,
graph{(9x^2)/25+(4y^2)/25=1 [-5, 5, -2.5, 2.5]}