The equation of the common tangent of the par | vedclass

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(B) The given parabolas are $x^2 = 108y$ (where $4a = 108 \implies a = 27$) and $y^2 = 32x$ (where $4b = 32 \implies b = 8$).The equation of a common

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$2x + 3y + 36 = 0$

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(B) The given parabolas are $x^2 = 108y$ (where $4a = 108 \implies a = 27$) and $y^2 = 32x$ (where $4b = 32 \implies b = 8$).The equation of a common tangent to parabolas $x^2 = 4ay$ and $y^2 = 4bx$ is given by $x(a^{1/3}) + y(b^{1/3}) + (a^{2/3}b^{2/3}) = 0$ is incorrect; the standard form is $x(b^{1/3}) + y(a^{1/3}) + (a^{1/3}b^{1/3}) = 0$ is not applicable here.Using the general formula for common tangent $x(a^{1/3}) + y(b^{1/3}) + (a^{2/3}b^{2/3}) = 0$ is for $y^2=4ax$ and $x^2=4by$. For $x^2 = 4Ay$ and $y^2 = 4Bx$,the common tangent is $x(B^{1/3}) + y(A^{1/3}) + (A^{1/3}B^{1/3}) = 0$ is not correct.Let the tangent to $x^2 = 108y$ be $y = mx - 27m^2$. Since this is also tangent to $y^2 = 32x$,we substitute $y = mx - 27m^2$ into $y^2 = 32x$:$(mx - 27m^2)^2 = 32x$$m^2x^2 - 54m^3x + 729m^4 - 32x = 0$$m^2x^2 - (54m^3 + 32)x + 729m^4 = 0$For tangency,the discriminant $D = 0$:$(54m^3 + 32)^2 - 4(m^2)(729m^4) = 0$$(54m^3 + 32)^2 - 2916m^6 = 0$$2916m^6 + 3456m^3 + 1024 - 2916m^6 = 0$$3456m^3 = -1024$$m^3 = -1024 / 3456 = -8 / 27$$m = -2/3$Substituting $m = -2/3$ into $y = mx - 27m^2$:$y = (-2/3)x - 27(-2/3)^2$$y = -2x/3 - 27(4/9)$$y = -2x/3 - 12$$3y = -2x - 36$$2x + 3y + 36 = 0$.

The equation of the common tangent of the parabolas $x^2 = 108y$ and $y^2 = 32x$ is

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