Question

If $x = a\left(t - \frac{1}{t}\right)$ and $y = b\left(t + \frac{1}{t}\right)$, then $\frac{dy}{dx} =$

• A. $\frac{a^2 x}{b^2 y}$
• B. $\frac{a^2 y}{b^2 x}$
• C. $-\frac{b^2 x}{a^2 y}$
• D. $\frac{b^2 x}{a^2 y}$

Hint:

Eliminating parameters using identities is an incredible shortcut for problems containing $(t + \frac{1}{t})$ and $(t - \frac{1}{t})$. Recognizing that it models a hyperbola of the form $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 4$ allows you to differentiate implicitly, completely skipping complex quotient rule work with parameter fractions!

Answer 1

The Correct Option is D

Step 1: Differentiate $x$ with respect to $t$.
$x = a\!\left(t - \dfrac{1}{t}\right)$, so $\dfrac{dx}{dt} = a\!\left(1 + \dfrac{1}{t^2}\right)$.
Step 2: Differentiate $y$ with respect to $t$.
$y = b\!\left(t + \dfrac{1}{t}\right)$, so $\dfrac{dy}{dt} = b\!\left(1 - \dfrac{1}{t^2}\right)$.
Step 3: Form $\dfrac{dy}{dx}$.
$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{b\!\left(1 - \frac{1}{t^2}\right)}{a\!\left(1 + \frac{1}{t^2}\right)} = \dfrac{b}{a}\cdot\dfrac{t^2 - 1}{t^2 + 1}$.
Step 4: Express the parameter ratio in $x, y$.
Note $\dfrac{x}{a} = t - \dfrac{1}{t} = \dfrac{t^2 - 1}{t}$ and $\dfrac{y}{b} = t + \dfrac{1}{t} = \dfrac{t^2 + 1}{t}$. Dividing, $\dfrac{x/a}{y/b} = \dfrac{t^2 - 1}{t^2 + 1}$.
Step 5: Substitute back.
So $\dfrac{dy}{dx} = \dfrac{b}{a}\cdot\dfrac{x/a}{y/b} = \dfrac{b}{a}\cdot\dfrac{bx}{ay}$.
Step 6: Simplify.
$\dfrac{dy}{dx} = \dfrac{b^2 x}{a^2 y}$.
\[ \boxed{\dfrac{dy}{dx} = \dfrac{b^2 x}{a^2 y}} \]