The curve $C$ has cartesian equation $y = \sqrt{4-2x^2}$. Sketch the curve $C$, labelling the

Question

The curve $C$ has cartesian equation $y = \sqrt{4-2x^2}$. Sketch the curve $C$, labelling the exact coordinates of any axial intercepts.[2] Find the equation of the tangent to $C$ at the point $P(1, \sqrt{2})$, leaving your answer in exact form.[2] Using the substitution $x = \sqrt{2} \sin \theta $, find the exact value of $\displaystyle\int_{1}^{\sqrt{2}} \sqrt{4-2x^2} \, \mathrm{d}x$.[4] Hence, find the exact value of the area of the region bounded by $C$, the tangent to $C$ at $P$, and the $x$-axis.[2]

Accepted Answer

(b)$y=-\sqrt 2x+ 2\sqrt 2$ (c)$\sqrt 2(\frac{\pi}{4}-\frac{1}{2})$ (d)$\sqrt 2-\frac{\sqrt 2}{4}\pi$ units$^2$

2025 RVHS P1 Q11

2025 RVHS P1 Q11