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

Oct 27, 2016

Critical points are $\left(0 , \frac{5}{2}\right)$, $\left(- \frac{5}{3} , 0\right)$, $\left(0 , - \frac{5}{2}\right)$, $\left(\frac{5}{3} , 0\right)$,

#### 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 $\frac{9 {x}^{2}}{25} + \frac{4 {y}^{2}}{25} = 1$, $\frac{\mathrm{dy}}{\mathrm{dx}}$ is given by

$\frac{9}{25} \times 2 x + \frac{4}{25} \times 2 y \times \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

i.e. $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{18 x}{25} \div \frac{8 y}{25} = - \frac{9 x}{4 y}$,

which is $0$, when $x = 0$ i.e. at $\frac{4 {y}^{2}}{25} = 1$ or $y = \pm \frac{5}{2}$

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