62\\a^2x^4+b^2y^4=c^6 \\a^2x^4+b^2y^4+2abx^2y^2=c^6+2abx^2y^2 \\(ax^2+by^2)^2-c^6=2abx^2y^2
Now can you try?
there is slightly moer than what I have written.. but i am sure that will suffice..
11Parametirc form of x=\frac{c^{\frac{3}{2}}\sqrt{sin\alpha}\sqrt[4]{2}}{\sqrt{a} } and that of y=\frac{c^{\frac{3}{2}}\sqrt{cos\alpha}\sqrt[4]{2}}{\sqrt{b} }.
So maximum is \frac{c^3}{\sqrt{ab}} if no mistake made.
Nishantda, can u pls complete ur line of thought?
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1Well , AM - GM would also do the job .
a 2 x 4 + b 2 y 42 ≥ √ a 2 b 2 x 4 y 4
or , c 62 ≥ a b x 2 y 2
So , c 62 a b ≥ x 2 y 2
So the minimum comes out to be ----- ( - c 3√ 2 a b ) And the maximum ----- ( c 3√ 2 a b )
62It is not given that a and b are positive..
that is why i used the squaring approach..
1@ Soumik, how did u get those parametric forms?
@ Nishant sir, sir can u pls complete ur proof?
1After what Nishant Sir has done -
By applying AM - GM inequality , one easily finds that ,
( p + q ) 2 ≥ 4 p q
Hence ,
c 6 + 2 a b x 2 y 2 = ( a x 2 + b y 2 ) 2 ≥ 4 a x 2 b y 2 = 4 a b x 2 y 2
So , c 6 + 2 a b x 2 y 2 - 4 a b x 2 y 2 ≥ 0
or , c 6 ≥ 2 a b x 2 y 2
or , c 62 a b ≥ x 2 y 2
From where we get the maximum value of " x y " as -
Max { x y } = c 3√2 a b
P . S - Every variable I have used is a positive integer except " c " .