The equation of the common tangent to the par | vedclass

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(A) For the parabola $y^2 = 4ax$,the tangent is $y = mx + \frac{a}{m}$. Here $4a = 32$,so $a = 8$. Thus,the tangent is $y = mx + \frac{8}{m}$.For the

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$x + 2y + 32 = 0$

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(A) For the parabola $y^2 = 4ax$,the tangent is $y = mx + \frac{a}{m}$. Here $4a = 32$,so $a = 8$. Thus,the tangent is $y = mx + \frac{8}{m}$.For the parabola $x^2 = 4by$,the tangent is $x = my - bm^2$. Here $4b = 256$,so $b = 64$. Thus,the tangent is $x = my - 64m^2$.Rewriting the first equation: $mx - y + \frac{8}{m} = 0$. Rewriting the second equation: $x - my + 64m^2 = 0$.Since these represent the same line,the ratio of coefficients must be equal: $\frac{m}{1} = \frac{-1}{-m} = \frac{8/m}{64m^2}$.From $\frac{m}{1} = \frac{1}{m}$,we get $m^2 = 1$,so $m = \pm 1$.Checking the ratio $\frac{8/m}{64m^2} = \frac{1}{8m^3}$.For $m = 1$,the ratio is $\frac{1}{8} 
eq 1$. For $m = -1$,the ratio is $\frac{1}{-8} 
eq -1$.Let the tangent be $y = mx + c$. For $y^2 = 32x$,$c = \frac{8}{m}$. For $x^2 = 256y$,$c = -64m^2$.Equating $c$: $\frac{8}{m} = -64m^2 \implies m^3 = -\frac{8}{64} = -\frac{1}{8} \implies m = -\frac{1}{2}$.Then $c = -64(-\frac{1}{2})^2 = -64(\frac{1}{4}) = -16$.The equation is $y = -\frac{1}{2}x - 16$,which simplifies to $2y = -x - 32$,or $x + 2y + 32 = 0$.

The equation of the common tangent to the parabolas $y^2 = 32x$ and $x^2 = 256y$ is

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