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

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Question

(A) The given equation is $6y = 2a^3x^2 + 3a^2x - 12a$. Dividing by $2a^3$ (assuming $a 
eq 0$), we get $x^2 + \frac{3}{2a}x = \frac{6y + 12a}{2a^3}$

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given equation is $6y = 2a^3x^2 + 3a^2x - 12a$. Dividing by $2a^3$ (assuming $a 
eq 0$), we get $x^2 + \frac{3}{2a}x = \frac{6y + 12a}{2a^3}$. Completing the square for $x$: $(x + \frac{3}{4a})^2 = \frac{6y + 12a}{2a^3} + \frac{9}{16a^2} = \frac{48y + 96a + 9a}{16a^3} = \frac{48y + 105a}{16a^3}$. Thus, $(x + \frac{3}{4a})^2 = \frac{3}{a^2}(y + \frac{35a}{16})$. The vertex $(h, k)$ is given by $h = -\frac{3}{4a}$ and $k = -\frac{35a}{16}$. From $h = -\frac{3}{4a}$, we have $a = -\frac{3}{4h}$. Substituting $a$ into $k$: $k = -\frac{35}{16} 	imes (-\frac{3}{4h}) = \frac{105}{64h}$. Therefore, $hk = \frac{105}{64}$. The locus is $xy = \frac{105}{64}$.

The locus of the vertices of the family of parabolas $6y = 2a^3x^2 + 3a^2x - 12a$ is

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