Q.If $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$, then show that $\dfrac{dy}{dx} \cdot \dfrac{dx}{dy} = 1$.

Concept: Implicit Differentiation & Inverse Function Rule

We treat $y$ as a function of $x$ and $x$ as a function of $y$ — the product of the derivatives must be $1$ for any smooth relation.

Step 1: Differentiate the given equation implicitly w.r.t. $x$:

Step 2: Factor out $\frac{dy}{dx}$:

Thus

Step 3: Differentiate the original equation implicitly w.r.t. $y$ (treat $x$ as function of $y$):

Factor $\frac{dx}{dy}$: …

The problem asks you to show that $\frac{dy}{dx} \cdot \frac{dx}{dy} = 1$ for the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. At first glance, this might look like a heavy algebraic exercise — but it’s actually a simple conceptual truth in calculus.

The key idea: if $y$ is a function of $x$ (implicitly defined by the equation), then $x$ is also a function of $y$ (locally, where the inverse exists). The derivatives $\frac{dy}{dx}$ and $\frac{dx}{dy}$ are reciprocals of each other. Their product is $1$ by definition — no matter how complicated the equation is.

Let’s verify this step by step.

1. Differentiate the given equation with respect to $x$.

   Treat $y$ as a function of $x$. Differentiate term by term:

   • $\frac{d}{dx}(ax^2) = 2ax$
   • $\frac{d}{dx}(2hxy) = 2h \left( y + x \frac{dy}{dx} \right)$ (product rule)
   • $\frac{d}{dx}(by^2) = 2by \frac{dy}{dx}$ (chain rule)
   • $\frac{d}{dx}(2gx) = 2g$
   • $\frac{d}{dx}(2fy) = 2f \frac{dy}{dx}$
   • $\frac{d}{dx}(c) = 0$

   Putting it together:

2. Collect terms containing $\frac{dy}{dx}$. Group the $\frac{dy}{dx}$ terms:

The remaining terms (without $\frac{dy}{dx}$) are:

So the equation becomes:

3. Solve for $\frac{dy}{dx}$. Divide through by 2:

Hence:

4. Now differentiate the same equation with respect to $y$.

   This time, treat $x$ as a function of $y$. Differentiate term by term:

   • $\frac{d}{dy}(ax^2) = 2ax \frac{dx}{dy}$
   • $\frac{d}{dy}(2hxy) = 2h \left( x\frac{dy}{dy} + y\frac{dx}{dy} \right) = 2h\left(x + y\frac{dx}{dy}\right)$
   • $\frac{d}{dy}(by^2) = 2by$
   • $\frac{d}{dy}(2gx) = 2g\frac{dx}{dy}$
   • $\frac{d}{dy}(2fy) = 2f$
   • $\frac{d}{dy}(c) = 0$

   Collecting:

5. Group $\frac{dx}{dy}$ terms. Terms with $\frac{dx}{dy}$:

Remaining terms:

So:

$2\frac{dx}{dy}(ax + hy + g) + 2(hx + by + f) = 0$ …

Method: The Reciprocal Relationship Between dy/dx and dx/dy

This method proves a general identity — that $\dfrac{dy}{dx}\cdot\dfrac{dx}{dy}=1$ for any relation connecting $x$ and $y$, as long as both derivatives exist and are non-zero. It does not depend on the specific form of the equation.

Steps

Step 1: Differentiate the given relation with respect to $x$, treating $y$ as a function of $x$

Apply implicit differentiation term by term (chain rule on every $y$-term, product rule on any mixed term) to obtain an equation you can solve for $\dfrac{dy}{dx}$ in terms of $x$ and $y$.

Step 2: Differentiate the same relation with respect to $y$, treating $x$ as a function of $y$

This mirrors Step 1: every $x$-term now picks up a factor of $\dfrac{dx}{dy}$ (chain rule), while every $y$-term differentiates directly. Solve this second equation for $\dfrac{dx}{dy}$.

Step 3: Multiply the two results

Because both expressions come from differentiating the same underlying relation — just with the roles of $x$ and $y$ swapped — the algebraic results from Steps 1 and 2 are exact reciprocals of each other, so their product collapses to $1$ once simplified. …

Common Mistakes

Mistake 1: Skipping the actual derivation and just asserting the reciprocal identity

Some students, knowing that $\frac{dy}{dx}$ and $\frac{dx}{dy}$ are "always reciprocals," skip differentiating the given conic equation entirely and just state the result. Why it's wrong: the question specifically asks you to show this using the given relation — an unsupported assertion earns no method marks even if the final claim is true in general. Correct approach: differentiate the equation once with respect to $x$ (to get $\frac{dy}{dx}$) and once with respect to $y$ (to get $\frac{dx}{dy}$), then multiply the two results.

Mistake 2: Not swapping roles correctly when differentiating with respect to $y$ …