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NCERT Exemplar · Q58

Q.If ax2+2hxy+by2+2gx+2fy+c=0ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0, then show that dydx⋅dxdy=1\dfrac{dy}{dx} \cdot \dfrac{dx}{dy} = 1.

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For any implicit relation, the derivative dydx\frac{dy}{dx} and its reciprocal dxdy\frac{dx}{dy} are multiplicative inverses — their product is always 11, provided neither derivative is zero or undefined. This follows directly from the chain rule and holds regardless of the specific equation.

The problem asks you to show that dydx⋅dxdy=1\frac{dy}{dx} \cdot \frac{dx}{dy} = 1 for the general second-degree equation ax2+2hxy+by2+2gx+2fy+c=0ax^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 yy is a function of xx (implicitly defined by the equation), then xx is also a function of yy (locally, where the inverse exists). The derivatives dydx\frac{dy}{dx} and dxdy\frac{dx}{dy} are reciprocals of each other. Their product is 11 by definition — no matter how complicated the equation is.

Let’s verify this step by step.

  1. Differentiate the given equation with respect to xx.

    Treat yy as a function of xx. Differentiate term by term:

    • ddx(ax2)=2ax\frac{d}{dx}(ax^2) = 2ax
    • ddx(2hxy)=2h(y+xdydx)\frac{d}{dx}(2hxy) = 2h \left( y + x \frac{dy}{dx} \right) (product rule)
    • ddx(by2)=2bydydx\frac{d}{dx}(by^2) = 2by \frac{dy}{dx} (chain rule)
    • ddx(2gx)=2g\frac{d}{dx}(2gx) = 2g
    • ddx(2fy)=2fdydx\frac{d}{dx}(2fy) = 2f \frac{dy}{dx}
    • ddx(c)=0\frac{d}{dx}(c) = 0

    Putting it together:

2ax+2h(y+xdydx)+2bydydx+2g+2fdydx=02ax + 2h\left(y + x\frac{dy}{dx}\right) + 2by\frac{dy}{dx} + 2g + 2f\frac{dy}{dx} = 0

  1. Collect terms containing dydx\frac{dy}{dx}. Group the dydx\frac{dy}{dx} terms:

2hxdydx+2bydydx+2fdydx=2dydx(hx+by+f)2hx\frac{dy}{dx} + 2by\frac{dy}{dx} + 2f\frac{dy}{dx} = 2\frac{dy}{dx}(hx + by + f)

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

2ax+2hy+2g2ax + 2hy + 2g

So the equation becomes:

2(ax+hy+g)+2dydx(hx+by+f)=02(ax + hy + g) + 2\frac{dy}{dx}(hx + by + f) = 0

  1. Solve for dydx\frac{dy}{dx}. Divide through by 2:

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

Hence:

dydx=−ax+hy+ghx+by+f\frac{dy}{dx} = -\frac{ax + hy + g}{hx + by + f}

  1. Now differentiate the same equation with respect to yy.

    This time, treat xx as a function of yy. Differentiate term by term:

    • ddy(ax2)=2axdxdy\frac{d}{dy}(ax^2) = 2ax \frac{dx}{dy}
    • ddy(2hxy)=2h(xdydy+ydxdy)=2h(x+ydxdy)\frac{d}{dy}(2hxy) = 2h \left( x\frac{dy}{dy} + y\frac{dx}{dy} \right) = 2h\left(x + y\frac{dx}{dy}\right)
    • ddy(by2)=2by\frac{d}{dy}(by^2) = 2by
    • ddy(2gx)=2gdxdy\frac{d}{dy}(2gx) = 2g\frac{dx}{dy}
    • ddy(2fy)=2f\frac{d}{dy}(2fy) = 2f
    • ddy(c)=0\frac{d}{dy}(c) = 0

    Collecting:

2axdxdy+2h(x+ydxdy)+2by+2gdxdy+2f=02ax\frac{dx}{dy} + 2h\left(x + y\frac{dx}{dy}\right) + 2by + 2g\frac{dx}{dy} + 2f = 0

  1. Group dxdy\frac{dx}{dy} terms. Terms with dxdy\frac{dx}{dy}:

2axdxdy+2hydxdy+2gdxdy=2dxdy(ax+hy+g)2ax\frac{dx}{dy} + 2hy\frac{dx}{dy} + 2g\frac{dx}{dy} = 2\frac{dx}{dy}(ax + hy + g)

Remaining terms:

2hx+2by+2f2hx + 2by + 2f

So:

2dxdy(ax+hy+g)+2(hx+by+f)=02\frac{dx}{dy}(ax + hy + g) + 2(hx + by + f) = 0 …

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