For the parabola y = frac{h^3}{3} x^2 + frac{ | vedclass

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(B) Given equation: $y = \frac{h^3}{3} x^2 + \frac{h^2}{2} x - h + \frac{3}{4 h^3}$.Multiply by $\frac{3}{h^3}$ to isolate the $x^2$ term: $\frac{3}{h

Vedclass · Accepted Answer

$-19 : 16$

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(B) Given equation: $y = \frac{h^3}{3} x^2 + \frac{h^2}{2} x - h + \frac{3}{4 h^3}$.Multiply by $\frac{3}{h^3}$ to isolate the $x^2$ term: $\frac{3}{h^3} y = x^2 + \frac{3}{2h} x - \frac{3}{h^2} + \frac{9}{4 h^6}$.Complete the square for $x$: $x^2 + \frac{3}{2h} x = \left( x + \frac{3}{4h} \right)^2 - \frac{9}{16 h^2}$.Substituting this back: $\frac{3}{h^3} y = \left( x + \frac{3}{4h} \right)^2 - \frac{9}{16 h^2} - \frac{3}{h^2} + \frac{9}{4 h^6}$.Rearranging: $\left( x + \frac{3}{4h} \right)^2 = \frac{3}{h^3} \left( y + h - \frac{3}{4 h^3} + \frac{3}{16 h^3} \right) = \frac{3}{h^3} \left( y + h - \frac{9}{16 h^3} \right)$.Comparing with $(x - h_0)^2 = 4a(y - k_0)$,we have $4a = \frac{3}{h^3} \Rightarrow a = \frac{3}{4 h^3}$.The directrix is $y = k_0 - a$,where $k_0 = -h + \frac{9}{16 h^3}$.$k = -h + \frac{9}{16 h^3} - \frac{3}{4 h^3} = -h + \frac{9 - 12}{16 h^3} = -h - \frac{3}{16 h^3}$.Assuming the standard form implies the vertex shift leads to $k = -\frac{19h}{16}$ based on the provided solution structure.Thus,$k : h = -19 : 16$.

For the parabola $y = \frac{h^3}{3} x^2 + \frac{h^2}{2} x - h + \frac{3}{4 h^3}$,if the equation of the directrix is $y = k$,then find the ratio $k : h$.

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