Question

If $a{x}^2+2hxy+b{y}^2=1,$ then $\frac{d^{2}y}{d{x}^2}$ is equal to

• A. $\frac{h^2-ab}{{\left(hx+by\right)}^2}$
• B. $\frac{h^2-ab}{{\left(hx-by\right)}^3}$
• C. $\frac{h^2-ab}{{\left(hx+by\right)}^3}$
• D. none of these.

Answer 1

Answer: (B)

$a{x}^2+2hxy+b{y}^2=1$
$\Rightarrow 2ax+2h\left(x\frac{dy}{dx}+y.1\right)+2by\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=-\frac{ax+hy}{hx+by}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}$
$=-\frac{\left(hx+by\right)\left(a+h\frac{dy}{dx}\right)-\left(ax+hy\right)\left(h+b\frac{dy}{dx}\right)}{{\left(hx+by\right)}^2}$
$=-\frac{\left(ahx+aby-ahx-h^{2}y\right)+\left(h^{2}x+hby-abx-hby\right)\frac{dy}{dx}}{{\left(hx+by\right)}^2}$
$=-\frac{y\left(ab-h^2\right)+x\left(h^2-ab\right)\frac{dy}{dx}}{{\left(hx+by\right)}^2}$
$=-\frac{y\left(ab-h^2\right)+x\left(h^2-ab\right)-\frac{ax+hy}{hx+by}}{{\left(hx+by\right)}^2}$
$=-\frac{\left(hyx+b{y}^2\right)\left(ab-h^2\right)-\left(h^2-ab\right)\left(a{x}^2+bxy\right)}{{\left(hx+by\right)}^3}$
$=-\frac{\left(ab-h^2\right)\left(hxy+b{y}^2+a{x}^2+hxy\right)}{{\left(hx+by\right)}^3}$
$=-\frac{\left(ab-h^2\right)\left(a{x}^2+2hxy+b{y}^2\right)}{{\left(hx+by\right)}^3}$
$=\frac{h^2-ab}{{\left(hx+by\right)}^3}\left[\because a{x}^2+2hxy+b{y}^2=1\right]$