Question

If $ax^{2} + 2 hxy + by^{2} = 1 ,$ then $ \frac{d^{2}y}{dx^{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)

$ax^{2} + 2hxy +by^{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}{dx^{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}} $
$= - \frac{\left(hyx +by^{2}\right)\left(ab-h^{2} \right) - \left(h^{2 } -ab\right)\left(ax^{2} +bxy\right)}{\left(hx +by\right)^{3}}$
$ = - \frac{\left(ab -h^{2}\right)\left(hxy +by^{2} + ax^{2} +hxy\right)}{\left(hx +by\right)^{3}}$
$ = - \frac{\left(ab -h^{2}\right)\left(ax^{2} + 2hxy + by^{2}\right)}{\left(hx +by\right)^{3}} $
$= \frac{h^{2} -ab}{\left(hx+by\right)^{3}} \left[\because ax^{2} + 2hxy +by^{2} = 1 \right] $