The locus of the vertices of the family of pa | vedclass

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(C) The given equation of the parabola is $y = \frac{a^3 x^2}{3} + \frac{a^2 x}{2} - 2a$.To find the vertex,we complete the square for $x$:$y = \frac{

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$xy = \frac{105}{64}$

Vedclass · Answer

(C) The given equation of the parabola is $y = \frac{a^3 x^2}{3} + \frac{a^2 x}{2} - 2a$.To find the vertex,we complete the square for $x$:$y = \frac{a^3}{3} \left( x^2 + \frac{3}{2a} x \right) - 2a$$y = \frac{a^3}{3} \left( x^2 + \frac{3}{2a} x + \left(\frac{3}{4a}\right)^2 - \left(\frac{3}{4a}\right)^2 \right) - 2a$$y = \frac{a^3}{3} \left( x + \frac{3}{4a} \right)^2 - \frac{a^3}{3} \cdot \frac{9}{16a^2} - 2a$$y = \frac{a^3}{3} \left( x + \frac{3}{4a} \right)^2 - \frac{3a}{16} - 2a$$y + \frac{35a}{16} = \frac{a^3}{3} \left( x + \frac{3}{4a} \right)^2$.The vertex $(h, k)$ is given by $h = -\frac{3}{4a}$ and $k = -\frac{35a}{16}$.To find the locus,we eliminate $a$:From $h = -\frac{3}{4a}$,we get $a = -\frac{3}{4h}$.Substitute this into $k = -\frac{35a}{16}$:$k = -\frac{35}{16} \left( -\frac{3}{4h} \right) = \frac{105}{64h}$.Therefore,$hk = \frac{105}{64}$.Replacing $(h, k)$ with $(x, y)$,the locus is $xy = \frac{105}{64}$.

List-$I$ (Geometric Property)	List-$II$ (Coordinates/Equations)
$I$. Vertex	$A$. $\left(-\frac{3}{2}, -3\right)$
$II$. Focus	$B$. $\left(\frac{3}{2}, -3\right)$
$III$. Equation of the directrix	$C$. $2x + 5 = 0$
$IV$. Equation of the axis	$D$. $2x + y + 3 = 0$
	$E$. $y + 3 = 0$
	$F$. $(-2, -3)$

The locus of the vertices of the family of parabolas $y = \frac{a^3 x^2}{3} + \frac{a^2 x}{2} - 2a$ is

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