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Q.
The curve represented by $x=3(\cos t+\sin t), y=4(\cos t-\sin t)$, is $-$
Conic Sections
Solution:
Given $\frac{x}{3}=\cos t+\sin t \& \frac{y}{4}=\cos t-\sin t$
Squaring these two,
$\Rightarrow \frac{x^2}{9}=1+2 \cos t \sin t $.........(i)
$\frac{y^2}{16}=1-2 \sin t \cos t$........(ii)
Adding (i) & (ii)
$\frac{x^2}{9}+\frac{y^2}{16}=2 \Rightarrow \frac{x^2}{18}+\frac{y^2}{32}=1$