The maximum number of common normals of y^2=4 | vedclass

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(C) The equations of the parabolas are $y^2=4ax$ and $x^2=4by$.The equation of the normal to $y^2=4ax$ with slope $m$ is $y=mx-2am-am^3$.The equation

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$5$

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(C) The equations of the parabolas are $y^2=4ax$ and $x^2=4by$.The equation of the normal to $y^2=4ax$ with slope $m$ is $y=mx-2am-am^3$.The equation of the normal to $x^2=4by$ with slope $m$ is $y=mx+2b+\frac{b}{m^2}$.For a common normal,the equations must be identical,so $-2am-am^3 = 2b+\frac{b}{m^2}$.Multiplying by $m^2$,we get $-2am^3-am^5 = 2bm^2+b$.Rearranging the terms,we obtain $am^5+2am^3+2bm^2+b=0$.Since this is a polynomial equation of degree $5$ in $m$,there are at most $5$ real roots for $m$.Therefore,the maximum number of common normals is $5$.

The maximum number of common normals of $y^2=4ax$ and $x^2=4by$ is equal to

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