A) A constant
B) A function of x only
C) A function of y only
D) A function of x and y
Correct Answer: A
Solution :
\[{{y}^{2}}=a{{x}^{2}}+bx+c\Rightarrow 2y\frac{dy}{dx}=2ax+b\] Þ \[2{{\left( \frac{dy}{dx} \right)}^{2}}+2y\frac{{{d}^{2}}y}{d{{x}^{2}}}=2a\Rightarrow y\frac{{{d}^{2}}y}{d{{x}^{2}}}=a-{{\left( \frac{dy}{dx} \right)}^{2}}\] Þ\[y\frac{{{d}^{2}}y}{d{{x}^{2}}}=a-{{\left( \frac{2ax+b}{2y} \right)}^{2}}\]Þ \[y\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{4a{{y}^{2}}-{{(2ax+b)}^{2}}}{4{{y}^{2}}}\] Þ \[4{{y}^{3}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=4a(a{{x}^{2}}+bx+c)-(4{{a}^{2}}{{x}^{2}}+4abx+{{b}^{2}})\] Þ\[4{{y}^{3}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=4ac-{{b}^{2}}\Rightarrow {{y}^{3}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{4ac-{{b}^{2}}}{4}=\]a constant.
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