If $ax^2 + 2hxy + by^2 = 0$ (here $a, b, h$ are real constants), then find $\frac{dy}{dx}$. Question:
If $ax^2 + 2hxy + by^2 = 0$ (Where $a, b, h$ are real constants), then find $\dfrac{dy}{dx}$.
Following choices are given:-

*

*$\dfrac yx$

*$\dfrac xy$

*$\dfrac {-y}x$

*$\dfrac {-x}y$

My work:
Differentiating the equation given,
$$2ax + 2h \left[y + x \dfrac{dy}{dx} \right] + 2by \dfrac{dy}{dx} = 0 $$
$$\implies \dfrac{dy}{dx} = \dfrac{-(ax+ hy)}{(hx+ by)}$$
Although I obtained $\dfrac{dy}{dx}$, but there is no such option given. I need to write the answer is terms of $x$ and $y$ only.
I tried to find the value of $h$ from the given equation and substituted in the value of $\dfrac{dy}{dx}$ but that seems not working here. What would be the appropriate way to solve this question?
 A: Method 1:
From the given equation, we have:
$$ax^2 + 2hxy + by^2 = 0 $$
$$\iff ax^2 + hxy + hxy + by^2 = 0 $$
$$\iff x(ax + hy) + y(hx + by) = 0 $$
$$\implies -\dfrac{ax + hy}{hx + by} = \dfrac{y}{x} $$
Thus, $$\dfrac{dy}{dx} =-\dfrac{ax + hy}{hx + by} = \dfrac{y}{x} $$
Hence option [A] is correct.

Alternative thinking:
$ax^2 + 2hxy + by^2 = 0 $ is a combined equation of two straight lines passing through $(0, 0)$ and thus the equation can be written in the form,
$$(y - m_1 x) (y- m_2x) = 0$$
From here, we have:
$$m_1 = \dfrac{y}{x}\quad and \quad m_2 = \dfrac{y}{x}$$
Since $\dfrac{dy}{dx}$ is the slope of the tangent line of the curve, thus $\dfrac{dy}{dx} = m_1 = m_2 = \dfrac{y}{x}$.
A: Divide the original equation by $x^2$ to obtain $b(\frac{y}{x})^2+2h\frac{y}{x}+a=0$.  Its solution is of the form $y=mx$ where $m$ is a constant satisfying $bm^2+2hm+a=0$. So $\frac{dy}{dx}=m=\frac{y}{x}$. Hence the first option is correct.
A: dividing by $xy$, and setting $z = \frac{y}{x}$, we get:
$bz^2 + 2hz + a = 0 \implies z = \frac{-2h \pm \sqrt{4h^2 - 4a}}{2b}$
Since $z = \frac{y}{x} = c$ is a constant, $\dfrac{dy}{dx}$ is also equal to $c$ which is same as $\frac{y}{x}$.
