If x=t^2+1, y=2 a t, then d^2 y d x^2 at t=a is - Vidyadip

MCQ

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

A
$-\frac{1}{a}$
B
$-\frac{1}{2 a^2}$
C
$\frac{1}{2 \sigma^2}$
D
$0$

✓

Answer

Given, $x=t^2+1$ and $y=2 a t$
\[
\begin{array}{l}
\Rightarrow \frac{d x}{d t}=2 t \Rightarrow \frac{d y}{d t}=2 a \therefore \frac{d y}{d x}=\frac{a}{t} \\
\Rightarrow \frac{d^2 y}{d x^2}=\frac{-a}{t^2} \cdot \frac{d t}{d x}=\frac{-a}{2 t^3} \therefore\left(\frac{d^2 y}{d x^2}\right)_{a t t=a}=\frac{-a}{2 a^3}=\frac{-1}{2 a^2}
\end{array}
\]

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