If y2=ax2+bx+c, then y3d2ydx2 is

A
a constant
B
$a function of x only$
C
$a function of y only$
D
a function of x and y

Solution:

$\begin{array}{l} y^{2} = a x^{2} + b x + c \\ d i f f w . r . t . x \\ \Rightarrow 2 y \frac{d y}{d x} = 2 a x + b \Rightarrow \frac{d y}{d x} = \frac{2 a x + b}{2 y} - - - - (1) \\ d i f f a g a i n w . r . t . x \\ \Rightarrow 2 {(\frac{d y}{d x})}^{2} + 2 y \frac{d^{2} y}{d x^{2}} = 2 a \\ \Rightarrow y \frac{d^{2} y}{d x^{2}} = a - {(\frac{d y}{d x})}^{2} \\ \Rightarrow y \frac{d^{2} y}{d x^{2}} = a - {(\frac{2 a x + b}{2 y})}^{2} f r o m (1) \\ \Rightarrow y \frac{d^{2} y}{d x^{2}} = \frac{4 a y^{2} - (2 a x + b)^{2}}{4 y^{2}} \\ 4 y^{3} \frac{d^{2} y}{d x^{2}} = 4 a (y^{2} - a x^{2} - b x) - b^{2} \\ 4 y^{3} \frac{d^{2} y}{d x^{2}} = 4 a c - b^{2} is a constant (from given c= y^{2} - a x^{2} - b x) \end{array}$

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

Solution:

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