Given equation is,
f ( x , y , z , p , q ) = x ( y 2 + z 2 × q 2 ) − z y 2 p = x y 2 + x z 2 q 2 − z y 2 p f(x,y,z,p,q) = x(y^2+z^2×q^2)- zy^2p= xy^2+xz^2q^2- zy^2p f ( x , y , z , p , q ) = x ( y 2 + z 2 × q 2 ) − z y 2 p = x y 2 + x z 2 q 2 − z y 2 p
Since, Transformation is, T = z 2 2 T = \frac{z^2}{2} T = 2 z 2
Now,
z = 2 T z = \sqrt{2T} z = 2 T
p = ∂ z ∂ x = 2 1 2 T − 1 / 2 ∂ T ∂ x = 1 2 T ∂ T ∂ x = 1 2 T P p =\frac{\partial z}{\partial x} = \sqrt{2}\frac{1}{2} T^{-1/2}\frac{\partial T}{\partial x} =\frac{1}{\sqrt{2T}} \frac{\partial T}{\partial x} =\frac{1}{\sqrt{2T}} P p = ∂ x ∂ z = 2 2 1 T − 1/2 ∂ x ∂ T = 2 T 1 ∂ x ∂ T = 2 T 1 P
q = ∂ z ∂ y = 2 1 2 T − 1 / 2 ∂ T ∂ y = 1 2 T ∂ T ∂ y = 1 2 T Q q=\frac{\partial z}{\partial y} = \sqrt{2}\frac{1}{2} T^{-1/2}\frac{\partial T}{\partial y} =\frac{1}{\sqrt{2T}} \frac{\partial T}{\partial y} =\frac{1}{\sqrt{2T}} Q q = ∂ y ∂ z = 2 2 1 T − 1/2 ∂ y ∂ T = 2 T 1 ∂ y ∂ T = 2 T 1 Q
Putting value of z, p and q in the given equation,
x y 2 + x ( 2 T ) ( 1 2 T Q ) 2 − 2 T y 2 ( 1 2 T P ) = 0 xy^2+x(2T)(\frac{1}{\sqrt{2T}} Q)^2 -\sqrt{2T} y^2 (\frac{1}{\sqrt{2T}} P)=0 x y 2 + x ( 2 T ) ( 2 T 1 Q ) 2 − 2 T y 2 ( 2 T 1 P ) = 0
x y 2 + x ( 2 T ) ( 1 2 T Q 2 ) − y 2 ( P ) = 0 xy^2+x(2T)(\frac{1}{{2T}} Q^2) - y^2 ( P)=0 x y 2 + x ( 2 T ) ( 2 T 1 Q 2 ) − y 2 ( P ) = 0
x y 2 + x Q 2 − y 2 P = 0 xy^2+xQ^2 - y^2 P=0 x y 2 + x Q 2 − y 2 P = 0
y 2 ( x − P ) = − x Q 2 y^2(x-P) = -xQ^2 y 2 ( x − P ) = − x Q 2
y 2 Q 2 = x ( P − x ) \frac{y^2}{Q^2} = \frac{x}{(P-x)} Q 2 y 2 = ( P − x ) x
It can written as,
g ( Q , y ) = f ( P , x ) g(Q,y) = f(P,x) g ( Q , y ) = f ( P , x )
which is the required result.
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