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

1 Answer
Shwetank Mauria
Oct 27, 2016

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]}

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