Q105 / 160 · TS EAMCET 2019 (04 May Shift 2) · Mathematics

If $t$ is a parameter and $x=t+\frac{1}{t}, y=t-\frac{1}{t}$, then $\frac{d^2 y}{d x^2}=$

  • A
    $\frac{4 t}{\left(t^2-1\right)^2}$
  • B
    $\frac{t^2}{\left(t^2-1\right)}$
  • C
    $\frac{4 t^3}{\left(t^2-1\right)^3}$
  • Correct answer
    D
    $\frac{-4 t^3}{\left(t^2-1\right)^3}$

Solution

We have, $$ \begin{aligned} x & =t+\frac{1}{t}, y=t-\frac{1}{t} \\ \therefore \quad \frac{d x}{d t} & =1-\frac{1}{t^2}, \frac{d y}{d t}=1+\frac{1}{t^2} \\ \frac{d y}{d x} & =\frac{d y / d t}{d x / d t}=\frac{1+\frac{1}{t^2}}{1-\frac{1}{t^2}}=\frac{t^2+1}{t^2-1} \\ \Rightarrow \quad \frac{d^2 y}{d x^2} & =\frac{\left(t^2-1\right)(2 t)-\left(t^2+1\right)(2 t)}{\left(t^2-1\right)^2} \times \frac{d t}{d x} \\ \Rightarrow \quad \frac{d^2 y}{d x^2} & =\frac{2 t\left(t^2-1-t^2-1\right)}{\left(t^2-1\right)^2} \times \frac{t^2}{\left(t^2-1\right)} \\ \frac{d^2 y}{d x^2} & =\frac{-4 t^3}{\left(t^2-1\right)^3} \end{aligned} $$

Hence, the correct answer is option: 'D'.

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