Find the slope of the graph of $xy-4y^2=2$ at $(9,2)$

Question

Find the slope of the graph of $xy-4y^2=2$ at $(9,2)$

I will take the derivative with respect to $x$, being careful to use implicit differentiation when differentiating a $y$ term, solving for $\frac{dy}{dx}$ ($\frac{dy}{dx}$ is the slope!!) and then plugging in the point $(9,2)$

Solution:

$\frac{d}{dx}(xy-4y^2)=\frac{d}{dx}2=0$

$(\frac{d}{dx}(xy)-4\frac{d}{dx}y^2)=0$

$(\frac{d}{dx}(xy)-4(2y\frac{dy}{dx})=0$

I will use the product rule to evaluate $\frac{d}{dx}(xy)$:

$((\frac{d}{dx}(x)y+x(\frac{d}{dx}y))-8y\frac{dy}{dx}=0$

$((1)y+x((1)\frac{dy}{dx}))-8y\frac{dy}{dx}=0$

$(y+x\frac{dy}{dx})-8y\frac{dy}{dx}=0$

$y+(x-8y)\frac{dy}{dx}=0$

$(x-8y)\frac{dy}{dx}=-y$

$\frac{dy}{dx}=\frac{-y}{(x-8y)}$

Cool, so now I've taken the derivative, was careful to use implicit differentiation when necessary, and solved for $\frac{dy}{dx}$, which tells us the slope at any $(x,y)$ point that we plug in. The problem statements wants us to find the slope at $(9,2)$, so let's plug that in!!

$\frac{dy}{dx}=\frac{-2}{(9-8(2))}$

$\frac{dy}{dx}=\frac{-2}{9-16}$

$\frac{dy}{dx}=\frac{-2}{-7}$

$\frac{dy}{dx}=\frac{2}{7}$

Answer 1

An other approach. The point $(9,2)$ belongs to $$4y^2-xy+2=0$$ $$\implies y=\frac{x+\sqrt{x^2-32}}{8}$$

the slope is then given by

$$\frac{dy}{dx}=\frac 18\left(1+ \frac{x}{\sqrt{x^2-32}}\right)$$

If $x=9$ then

$$\frac{dy}{dx}=\frac 18\left(1+ \frac 97\right)=\frac 27$$

Answer 2

Here is a quick calculation. Write $x = \frac 2y + 4y$ and $\frac{dx}{dy}=-\frac2{y^2}+4$. Then, plug in $y=2$ to obtain,

$$\frac{dy}{dx}=\frac1{\frac{dx}{dy}}=\frac1{-\frac2{2^2}+4}=\frac27$$

Answer 3

Looks good to me. You can verify the result by taking advantage of the fact that you can obtain an equation for the tangent to a conic at a point on the conic mechanically via a set of simple substitutions. Here we have $$xy-4y^2 = 2 \to \frac12(2x+9y)-4\cdot2y=2 \to x-\frac72y=2,$$ from which the slope of the line is $\frac27$, just as you calculated.