If $f (2 x + 3 y, 2 x - 3 y) = 24 x y$ then $f (x, y)$ is

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Solution

Finding the value of $f (x, y)$ :
Given that $f (2 x + 3 y, 2 x - 3 y) = 24 x y$
Let $2 x + 3 y = h & 2 x - 3 y = k$
$x = \frac{(h + k)}{4} & y = \frac{(h - k)}{6}$
$f (h, k) = 24 \times \frac{(h + k)}{4} \times \frac{(h - k)}{6}$
$\Rightarrow$ $= (h + k) (h - k)$
$\Rightarrow$ $f (h, k) = h^{2} - k^{2}$
Similarly $f (x, y) = x^{2} - y^{2}$
Hence, the value of $f (x, y)$ is $x^{2} - y^{2}$

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If $f (2 x + 3 y, 2 x - 3 y) = 24 x y$ then $f (x, y)$ is