A curve has equation $x^{\frac{1}{2}} + y^{\frac{1}{2}} = 3$, for $x 0$ and \(y 0

Question

A curve has equation $x^{\frac{1}{2}} + y^{\frac{1}{2}} = 3$, for $x > 0$ and $y > 0$. Show that $\frac{\mathrm{d}y}{\mathrm{d}x} = -\left(\frac{y}{x}\right)^{\frac{1}{2}}$.[2] Find the equation of the normal to the curve at the point where $x = 1$ and determine the acute angle that the normal makes with the $x$-axis.[5]

Accepted Answer

$\textbf{(b)} \quad y = \frac{1}{2}x + \frac{7}{2} \;;\; \tan^{-1}\frac{1}{2} \text{ or } 0.464 \text{ radians} $

N2021 P1 Q3 [Modified]

N2021 P1 Q3 [Modified]

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