MCQ
If $x=t^2+1, y=2 a t$, then $\frac{d^2 y}{d x^2}$ at $t=a$ is
  • A
    $-\frac{1}{a}$
  • $-\frac{1}{2 a^2}$
  • C
    $\frac{1}{2 \sigma^2}$
  • D
    $0$

Answer

Correct option: B.
$-\frac{1}{2 a^2}$
Given, $x=t^2+1$ and $y=2 a t$
$\Rightarrow \frac{d x}{d t}=2 t$
$\Rightarrow \frac{d y}{d t}=2 a$
$\therefore \frac{d y}{d x}=\frac{a}{t}$
$\Rightarrow \frac{d^2 y}{d x^2}=\frac{-a}{t^2} \cdot \frac{d t}{d x}=\frac{-a}{2 t^3}$
$\therefore\left(\frac{d^2 y}{d x^2}\right)_{a t t=a}=\frac{-a}{2 a^3}=\frac{-1}{2 a^2}$

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