Question

If $y^2=a{x}^2+bx+c$, then $\frac{d}{dx}\left(y^3{y}_2\right)=$

• A. $1$
• B. $-1$
• C. $\frac{4ac-b^2}{a^2}$
• D. $0$

Answer 1

Answer: (D)

Given $y^2=a{x}^2+bx+c$
Differentiating both sides, we get $2y{y}_1=2ax+b\dots (i)$
Again differentiating, we get $2y{y}_2+y_1(2y_1)=2a$
$\Rightarrow y{y}_2=a-y_1^2$
$\Rightarrow y{y}_2=a-{\left(\frac{2ax+b}{2y}\right)}^2$ (using $(i)$)
$=\frac{4y^{2}a-\left(4a^2{x}^2+b^2+4abx\right)}{4y^2}$
$\Rightarrow y^3{y}_2=\frac{4a\left(a{x}^2+bx+c\right)-\left(4a^2{x}^2+b^2+4abx\right)}{4}$
$=\frac{4ac-b^2}{4}$
$\Rightarrow \frac{d}{dx}\left(y^3{y}_2\right)=0$