Question

If $a{x}^2+2hxy+b{y}^2=0,$ then $\frac{dy}{dx}$ is

• A. $\frac{y}{x}$
• B. $\frac{x}{y}$
• C. $-\frac{x}{y}$
• D. None of these

Answer 1

The correct option is A $\frac{y}{x}$

${ax}^2+2hxy+{by}^2=0$

Differentiating w.r.t ${}^{\prime }{x}^{\prime }$

$2ax+2hy+2hx\frac{dy}{dx}+2by\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}(hx+by)=-ax-hy$

$\Rightarrow \frac{dy}{dx}=\frac{-(ax+hy)}{hx+by}⟶\left(1\right)$

We have

$\Rightarrow {ax}^2+2hxy+{by}^2=0$

$\Rightarrow {ax}^2+hxy+hxy+{by}^2=0$

$\Rightarrow x(ax+hy)+y(hx+by)=0$

$\Rightarrow \frac{x}{y}=\frac{-(hx+by)}{(ax+hy)}⟶\left(3\right)$

$\therefore$ Using $\left(2\right)$ we can write $\left(1\right)$ as

$\frac{dy}{dx}=\frac{y}{x}$

$\therefore$ Option A.