Find the equation of the common tangent to th | vedclass

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(B) The equation of any tangent to the parabola $x^2 = 16y$ is of the form $y = mx - 4m^2$,where $a = 4$ in $x^2 = 4ay$.If this line is also a tangent

Vedclass · Accepted Answer

$x + 2y + 2 = 0$

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(B) The equation of any tangent to the parabola $x^2 = 16y$ is of the form $y = mx - 4m^2$,where $a = 4$ in $x^2 = 4ay$.If this line is also a tangent to the parabola $y^2 = 2x$ (where $a = 1/2$),the condition for tangency $c = a/m$ must be satisfied.Here,$c = -4m^2$ and $a = 1/2$,so $-4m^2 = (1/2)/m$.This simplifies to $m^3 = -1/8$,which gives $m = -1/2$.Substituting $m = -1/2$ into the tangent equation: $y = (-1/2)x - 4(-1/2)^2$.$y = -x/2 - 1$.Multiplying by $2$,we get $2y = -x - 2$,which rearranges to $x + 2y + 2 = 0$.

Find the equation of the common tangent to the parabolas $y^2 = 2x$ and $x^2 = 16y$.

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