If {y^2} = a{x^2} + bx + c,then {y^3}frac{{{d | vedclass

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(A) Given ${y^2} = a{x^2} + bx + c$. Differentiating with respect to $x$,we get: $2y\frac{{dy}}{{dx}} = 2ax + b \Rightarrow \frac{{dy}}{{dx}} = \frac{

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$A$ constant

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(A) Given ${y^2} = a{x^2} + bx + c$. Differentiating with respect to $x$,we get: $2y\frac{{dy}}{{dx}} = 2ax + b \Rightarrow \frac{{dy}}{{dx}} = \frac{{2ax + b}}{{2y}}$. Differentiating again with respect to $x$: $2{\left( {\frac{{dy}}{{dx}}} \right)^2} + 2y\frac{{{d^2}y}}{{d{x^2}}} = 2a$. Dividing by $2$: ${\left( {\frac{{dy}}{{dx}}} \right)^2} + y\frac{{{d^2}y}}{{d{x^2}}} = a$. Substituting $\frac{{dy}}{{dx}} = \frac{{2ax + b}}{{2y}}$: $y\frac{{{d^2}y}}{{d{x^2}}} = a - {\left( {\frac{{2ax + b}}{{2y}}} \right)^2} = a - \frac{{{{(2ax + b)}^2}}}{{4{y^2}}}$. $y\frac{{{d^2}y}}{{d{x^2}}} = \frac{{4a{y^2} - {{(2ax + b)}^2}}}{{4{y^2}}}$. Substituting ${y^2} = a{x^2} + bx + c$: $4{y^3}\frac{{{d^2}y}}{{d{x^2}}} = 4a(a{x^2} + bx + c) - (4{a^2}{x^2} + 4abx + {b^2})$. $4{y^3}\frac{{{d^2}y}}{{d{x^2}}} = 4{a^2}{x^2} + 4abx + 4ac - 4{a^2}{x^2} - 4abx - {b^2} = 4ac - {b^2}$. Thus,${y^3}\frac{{{d^2}y}}{{d{x^2}}} = \frac{{4ac - {b^2}}}{4}$,which is a constant.

If ${y^2} = a{x^2} + bx + c$,then ${y^3}\frac{{{d^2}y}}{{d{x^2}}}$ is

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