A) 1
B) -1
C) \[\frac{4ac-{{b}^{2}}}{{{a}^{2}}}\]
D) 0
Correct Answer: D
Solution :
| Given \[{{y}^{2}}=a{{x}^{2}}+bx+c\] |
| \[\Rightarrow \,2y{{y}_{1}}=2ax+b\] (i) |
| \[\Rightarrow 2y{{y}_{2}}+{{y}_{1}}\left( 2{{y}_{1}} \right)=2a\] |
| \[\Rightarrow y{{y}_{2}}=a-y_{1}^{2}\] |
| \[\Rightarrow y{{y}_{2}}=a-{{\left( \frac{2ax+b}{2y} \right)}^{2}}\] (Using (i)) |
| \[=\frac{4{{y}^{2}}a-\left( 4{{a}^{2}}{{x}^{2}}+{{b}^{2}}+4abx \right)}{4{{y}^{2}}}\] |
| \[\Rightarrow \,{{y}^{3}}{{y}_{2}}=\frac{4a\left( a{{x}^{2}}+bx+c \right)-\left( 4{{a}^{2}}{{x}^{2}}+{{b}^{2}}+4abx \right)}{4}\]\[=\frac{4ac-{{b}^{2}}}{4}\] |
| \[\Rightarrow \frac{d}{dx}\left( {{y}^{3}}{{y}_{2}} \right)=0\] |
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