*x* ∝ *y ^{2}*and

*y*= 2a when

*x*=

*a*, then show that

*y*= 4

^{2}*ax*.

Solution : * x* ∝ *y ^{2}*

⇒ *x* = k*y ^{2}* ——————-( 1 )

where k is variation constant .

when *x* = a and *y* 2a, then from equation ( 1 ),

*a* = k ( 2a )^{2}

⇒ a = k × 4a^{2}

⇒ 1 = k ( 4a )

⇒ k = 1 / ( 4a )

putting the value of k = 1 / ( 4a ) in equation ( 1 ), we get,

*x* = k*y ^{2}*

⇒ *x* = { 1 / ( 4a ) } *y ^{2}*

∴ *y ^{2}* = 4a

*x*( showed )

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