The Y-intercept of the common tangent to the | vedclass

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

Vedclass · Accepted Answer

$-12$

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(B) The given parabolas are $y^2 = 32x$ (where $4a = 32 \implies a = 8$) and $x^2 = 108y$ (where $4b = 108 \implies b = 27$).Any tangent to $y^2 = 32x$ is of the form $y = mx + \frac{8}{m}$.This line is also a tangent to $x^2 = 108y$. Substituting $y = mx + \frac{8}{m}$ into $x^2 = 108y$ gives $x^2 = 108(mx + \frac{8}{m}) \implies x^2 - 108mx - \frac{864}{m} = 0$.Since the line is tangent,the discriminant $D = 0$:$(-108m)^2 - 4(1)(-\frac{864}{m}) = 0 \implies 11664m^2 + \frac{3456}{m} = 0$.$11664m^3 = -3456 \implies m^3 = -\frac{3456}{11664} = -\frac{1}{3.375} = -\frac{8}{27}$.Thus,$m = -\frac{2}{3}$.The $Y$-intercept is $c = \frac{8}{m} = \frac{8}{-2/3} = 8 	imes (-\frac{3}{2}) = -12$.

The $Y$-intercept of the common tangent to the parabola $y^2 = 32x$ and $x^2 = 108y$ is

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