Find dy/dx given x = a cos²phi, y = b sin²phi
Find $\dfrac{dy}{dx}$ given $x = a\cos^{2}\varphi$, $y = b\sin^{2}\varphi$.
Explanation of Method
A parametrisation whose image is a straight segment.
Step 1: Differentiate with respect to $\varphi$
Both derivatives produce the same double-angle factor: \[x'_{\varphi} = 2a\cos\varphi(-\sin\varphi) = -a\sin 2\varphi, \qquad y'_{\varphi} = 2b\sin\varphi\cos\varphi = b\sin 2\varphi.\]
Step 2: Form the ratio
The factor $\sin 2\varphi$ cancels completely, leaving a constant: $\frac{dy}{dx} = \frac{b\sin 2\varphi}{-a\sin 2\varphi} = -\frac{b}{a}$.
Step 3: Verification
A constant slope signals a straight line, and indeed the Pythagorean identity eliminates the parameter at once: $\frac{x}{a} + \frac{y}{b} = \cos^{2}\varphi + \sin^{2}\varphi = 1$.
\[\text{The line } \frac{x}{a}+\frac{y}{b} = 1 \text{ has slope } -\frac{b}{a} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\dfrac{dy}{dx} = -\dfrac{b}{a}, \qquad \sin 2\varphi \neq 0\,}\]
Remark
Remark. The image is not the whole line but only the segment joining $(a,0)$ and $(0,b)$, traced back and forth as $\varphi$ varies. The excluded values $\varphi = \tfrac{k\pi}{2}$ are the endpoints, where both parametric derivatives vanish and the point reverses direction — the parametrisation is singular there even though the curve itself is perfectly smooth.