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

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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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