Question

If \(x=at^{2}\) and \(y=2at,t\neq 0\) then ...

• A. \(x\frac{d^{2}x}{dy^{2}}+a=0\)
• B. \(2a\frac{d^{2}x}{dy^{2}}-1=0\)
• C. \(\frac{d^{2}x}{dy^{2}}-2a=0\)
• D. \(\frac{d^{2}x}{dy^{2}}-1=0\)

Hint:

Differentiate x with respect to y twice, using the parametric form with t.

Answer 1

The Correct Option is B

Step 1: Eliminate t
From $y=2at$, $t=\dfrac{y}{2a}$, so $x=a\cdot \dfrac{y^{2}}{4a^{2}}=\dfrac{y^{2}}{4a}$.

Step 2: Differentiate twice
$\dfrac{dx}{dy}=\dfrac{y}{2a}$ and $\dfrac{d^{2}x}{dy^{2}}=\dfrac{1}{2a}$.

Step 3: Match
Rearranging gives $2a\dfrac{d^{2}x}{dy^{2}}-1=0$.

Final Answer:
Direct elimination also gives $2a\dfrac{d^{2}x}{dy^{2}}=1$. \[ \boxed{\text{(B) }2a\dfrac{d^{2}x}{dy^{2}}-1=0} \]