I wanted to ask a question about using implicit differentiation to find coordinates where a normal to a hyperbola intersects at 2 different locations.
I'm studying fundamental advanced mathematics so I apologise if the error in my reasoning seems obvious!
I was presented with the following question:
The points $P(4, 12)$ and $Q(-8, -6)$ lie on the rectangular hyperbola $H$ with equation $xy = 48$.
The equation of the line $PQ$ is given to be $3x – 2y + 12 = 0$.
The point $A$ lies on $H$. The normal to $H$ at $A$ is parallel to the chord $PQ$.
Find the exact coordinates of the two possible positions of $A$
Given from the equation of the line, the gradient of the line $PQ$ is $\frac{3}{2}$ so the gradient of the tangent at $H$ is $-\frac{2}{3}$.
I then attempted implicit differentation of $xy = 48$:
$$x \frac{\mathrm{d}y}{\mathrm{d}x} + y = 0$$
and from here
$$-\frac{2}{3}x + y = 0$$
However, I am stuck here and not able to progress further.
In the solutions, the derivative of $y = \frac{48}{x}$ was taken and evaulated:
$$\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{48}{x^2} = -\frac{2}{3}$$
to yield
$$72 = x^2$$ and from there lead to the final solution.
The struggle that I'm facing is trying to understand why implicit differentiation isn't working to guide me to a final solution, unlike explicit differentiation.
I might have made an error or not realised the limitations of implicit differentiation in this context, but I'd like to know if there is a mistake that I have made or whether I've not considered the limitations of implicit differentation.
Best Answer
We have $$\dfrac{d}{dx}(xy)=\dfrac{d}{dx}(48)\Rightarrow y+x\frac{dy}{dx}=0\Rightarrow\frac{dy}{dx}=-\frac{y}{x}.$$ Thus, at the point $A(x_0,y_0)$ the slope of the tangent line is $-\dfrac{y_0}{x_0}$, therefore, the slope of the normal line will be given by $$\frac{x_0}{y_0}.$$ However, such line is parallel to the line $3x - 2y + 12 = 0\Leftrightarrow y=\dfrac{3}{2}x+6$, then $$\frac{x_0}{y_0}=\frac{3}{2}\Longrightarrow\left\{\begin{matrix}x_0&=3k\\y_0&=2k\end{matrix}\right., k\in\mathbb{R}.$$
But $A$ lies on the hyperbole $xy=48$, thus $$x_0y_0=48\Rightarrow 6k^2=48\Rightarrow k^2=8\Rightarrow k=\pm2\sqrt{2}.$$ Therefore, the possibles values for $A$ are $(6\sqrt{2},4\sqrt{2})$ and $(-6\sqrt{2},-4\sqrt{2})$.