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Q.
Maximum number of common normals of $y^{2}=4 a x$ and $x^{2}=4 b y$ may be equal to
Conic Sections
Solution:
Equations of normals to $y^{2}=4 a x$ and $x^{2}=4 b y$ are given by
$y=m x-2 a m-a m^{3}$ and $y=m x+2 b+\frac{b}{m^{2}}$
For common normals, $2 b+\frac{b}{m^{2}}+2 a m+a m^{3}=0$
$\Rightarrow a m^{5}+2 a m^{3}+2 b m^{2}+b=0$
So, a maximum of 5 normals are possible.