CBSE 2026 · Region 5 · Set 2 · Q34 · 5 marks
Sketch the curve described by $\displaystyle \left\{(\mathrm{x}, \mathrm{y}): 9 \mathrm{x}^{2}+16 \mathrm{y}^{2}=144\right\}$ and find the area of the region enclosed by it, using integration.
Marking-scheme solution
Required area $\displaystyle =4 \int_{0}^{4} \dfrac{3}{4} \sqrt{16-x^{2}}\, d x$
$\displaystyle =3\left[\dfrac{x}{2} \sqrt{16-x^{2}}+\dfrac{16}{2} \sin^{-1} \dfrac{x}{4}\right]_{0}^{4}$
$\displaystyle =12 \pi$
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.