Question

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

• A. a constant function
• B. a function of x only
• C. a function of y only
• D. a function of both x and y

Answer 1

The correct option is C a function of y only

$y^2=a{x}^2+bx+c$ is

Differentiating this equ. (i),

$2y\frac{dy}{dx}=2ax+6$

Differentiating above equation again,

$2y\frac{d^{2}y}{d{x}^2}+2\left(\frac{dy}{dx}\right)\left(\frac{dy}{dx}\right)=2a$

$\frac{d^{2}y}{d{x}^2}=\frac{2a-2{\left(\frac{dy}{dx}\right)}^2}{2y}=\frac{a-{\left(\frac{dy}{dx}\right)}^2}{y}$

$y^2\frac{{\alpha }^{2}y}{d{x}^2}=y^2\left(\frac{a-{\left(\frac{dy}{dx}\right)}^2}{y}\right)$

$y^2\frac{{\alpha }^{2}y}{d{x}^2}=y^2\left(\frac{a-{\left(\frac{dy}{dx}\right)}^2}{y}\right)$

$=y(a-{\left(\frac{dy}{dx}\right)}^2)\left\{\begin{array}{c}\because 2y\frac{dy}{dx}=2ax+b\\ \frac{dy}{dx}=\frac{2ax+b}{2y}\end{array}\right\}$

$=y(a-{\left(\frac{2ax+b}{2y}\right)}^2)$

$=\frac{y(4a{y}^2-(4a^2{x}^2+b^2+4abx)}{4y^2}$

$=\frac{4a{y}^2-4a^2{x}^2-b^2-4abx}{4y}$

$=\frac{4a(a{x}^2+6x+c)-4a^2{x}^2-6^2-4abx}{4y}$

$=\frac{4a^2{x}^2+4abx+4ac-4a^2{x}^2-b^2-4abx}{4y}$

$=\frac{4ac-b^2}{4y}$

$\therefore$ It is function of $y$ only.

option (C) is correct.