How to find all points on a curve

[pierre]

How to find all points on the curve xy-y^2=1 where the slope of the tangent line is -1/2

 

[mmm4444bot]

Use the Quadratic Formula to solve the given equation for y. You'll get two expressions in x.

Set the derivative of each equal to -1/2, and solve for x.

 

[soroban]

Hello, pierre!

$\displaystyle \text{Find all points on the curve }\:xy-y^2\:=\:1\;\;[1]$

. . $\displaystyle \text{where the slope of the tangent line is }\,\text{-}\tfrac{1}{2}$

$\displaystyle \text{Differentiate implicitly:}$

. . $\displaystyle x\tfrac{dy}{dx} + y - 2y\tfrac{dy}{dx} \:=\:0 \quad\Rightarrow\quad 2y\tfrac{dy}{dx} - x\tfrac{dy}{dx} \:=\:y \quad\Rightarrow\quad (2y-x)\tfrac{dy}{dx} \:=\:y$

. . $\displaystyle \text{Hence: }\;\frac{dy}{dx} \;=\;\frac{y}{2y-x}$


$\displaystyle \text{Then: }\;\frac{y}{2y-x} \;=\;-\frac{1}{2}$

. . $\displaystyle \text{Then: }\;2y \;=\;-2y + x \quad\Rightarrow\quad x \:=\:4y\;\;[2]$


$\displaystyle \text{Substitute into [1]: }\;(4y)y - y^2 \:=\:1 \quad\Rightarrow\quad 3y^2 \:=\:1 \quad\Rightarrow\quad y^2 \:=\:\frac{1}{3} \quad\Rightarrow\quad y \:=\:\pm\frac{1}{\sqrt{3}}$

$\displaystyle \text{Substitute into [2]: }\;x \:=\:4\left(\pm\frac{1}{\sqrt{3}}\right) \:=\:\pm\frac{4}{\sqrt{3}}$


$\displaystyle \text{Therefore: }\;\left(\frac{4}{\sqrt{3}},\;\frac{1}{\sqrt{3}}\right) \;\text{ and }\: \left(-\frac{4}{\sqrt{3}},\;-\frac{1}{\sqrt{3}}\right)$