Question

Consider the parametric equation $x=\frac{a(1-t^2)}{1+t^2},y=\frac{2at}{1+t^2}$.

What is $\frac{dy}{dx}$ equal to?

• A. $\frac{y}{x}$
• B. $-\frac{y}{x}$
• C. $\frac{x}{y}$
• D. $-\frac{x}{y}$

Answer 1

Answer: (D)

$-\frac{x}{y}$

Given : $x=\frac{a(1-t^2)}{1+t^2}andy=\frac{2at}{1+t^2}$
First consider $x=\frac{a(1-t^2)}{1+t^2}$
Diferentiating w.r.t ${}^{\prime }{t}^{\prime }$
$\frac{dx}{dt}=a\left[\frac{(1+t^2)(-2t)-\left(1{-t}^2\right)\left(2t\right)}{{(1+t^2)}^2}\right]$
$=-2at\left[\frac{(1+t^2+1-t^2)}{{(1+t^2)}^2}\right]=\frac{-4at}{{(1+t^2)}^2}\to \left(1\right)$
secondly $y=\frac{2at}{1+t^2}$
$\Rightarrow \frac{dy}{dt}=2a\left[\frac{(1+t^2.1)-t\left(2t\right)}{{(1+t^2)}^2}\right]\to \left(2\right)$
divide $\left(2\right)÷\left(1\right)$
$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2a\left[\frac{(1+t^2.1)-t\left(2t\right)}{{(1+t^2)}^2}\right]}{\frac{-4at}{{(1+t^2)}^2}}$
$=\frac{-1[1+t^2-2t^2]}{-4t}\frac{dy}{dx}=\frac{-1-t^2}{2t}\to \left(3\right)$
Given : $x=\frac{a(1-t^2)}{1+t^2}andy=\frac{2at}{1+t^2}$
$\frac{y}{x}=\frac{\frac{2at}{1+t^2}}{\frac{a(1-t^2)}{1+t^2}}$
$\frac{y}{x}=\frac{2t}{1-t^2}\to \left(4\right)$
Therefor substitute equation $\left(4\right)$ in $\left(3\right)$
$\frac{dy}{dx}=\frac{-x}{y}$
Hence option $\left(4\right)$ is the correct answer.