- #1

WMDhamnekar

MHB

- 364

- 28

$$\frac{(x+y)^2}{16}+\frac{(x-y)^2}{9}=1$$

Solution given:-

Let $x+y= 4\cos{\alpha},x-y=3\sin{\alpha}$ Then $x=\frac{4\cos{\alpha}+3\sin{\alpha}}{2}$ $\Rightarrow dx=\frac{3\cos{\alpha}-4\sin{\alpha}}{2}d\alpha$

$y=\frac{4\cos{\alpha}-3\sin{\alpha}}{2}$. So, $ydx=\left(3-\frac{25}{8}*\sin{2\alpha}\right)*d\alpha$

**What is this ydx?**

Hence the required area is

Hence the required area is

$$\displaystyle\int_0^{2\pi}\left(3-\frac{25}{8}*\sin{2\alpha}\right)*d\alpha$$

**Why does the author select ydx as integrand for computation of area of ellipse? What is the logic behind that?**

$=6\pi=18.849$

If any member of Math help Board knows the explanations for my queries, may reply to this question.