Answer is: option4
\( -2 \)Solution:
Solution I.
\( \frac{dx}{dt} = \cos t \) and \( \frac{dy}{dt} = -2\cos t \sin t \).
Hence \[ \frac{dy}{dx} = \frac{-2\cos t \sin t}{\cos t} = -2\sin t. \]
To compute \( \frac{d^2 y}{dx^2} \), we need
\[ \frac{d}{dt}\left(\frac{dy}{dx}\right) \Big/ \frac{dx}{dt} = \frac{-2\cos t}{\cos t} = -2. \]
The second derivative is constant, not depending on \( t = \frac{\pi}{2} \).
Solution II.
We eliminate the parameter. \( x^2 = \sin^2 t \) and \( y = \cos^2 t \), so \( x^2 + y = 1 \).
Hence \( y = 1 - x^2 \). Then \[ \frac{dy}{dx} = -2x \quad \text{and} \quad \frac{d^2 y}{dx^2} = -2. \]
