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Normal of a Curve y = f(x)

Trace the fol...

Question

Trace the following central conics.
$(3 x - 4 y + a) (4 x + 3 y + a) = a^{2} .$

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Solution

$(3 x - 4 y + a) (4 x + 3 y + a) = a^{2} (\frac{3 x - 4 y + a}{5}) (\frac{4 x + 3 y + a}{5}) = \frac{a^{2}}{25} X = \frac{3 x - 4 y + a}{5}, Y = \frac{4 x + 3 y + a}{5} X Y = {(\frac{a}{5})}^{2}$
Comparing with standard equation of rectangular hyperbola
$x y = c^{2} \Rightarrow c = \frac{a}{5}$

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