We have, the equation of the ellipse
x² + 9y² = 36
\(\Rightarrow \frac{x^2}{36}+\frac{y^2}{4}=1 \Rightarrow \frac{x^2}{6^2}+\frac{y^2}{2^2}=1\)

\(\therefore\) Required area \(=4 \int_0^6 y d x\)
\(\begin{aligned} & =4 \int_0^6 \frac{2}{6} \sqrt{36-x^2} d x=\frac{4}{3} \int_0^6 \sqrt{36-x^2} d x \end{aligned}\)
\(\begin{aligned} =\frac{4}{3}\left[\frac{x}{2} \sqrt{36-x^2}+\frac{36}{2} \sin ^{-1} \frac{x}{6}\right]_0^6 \end{aligned}\)
\(\begin{aligned} & =\frac{4}{3}\left[0+\frac{36}{2} \sin ^{-1} 1-0\right]=\frac{4}{3} \times 18 \times \frac{\pi}{2} \end{aligned}\)
= 12 \(\pi\) sq units

\(y=\sqrt{a^{2}-x^{2}}\)
Differentiating both sides of this equation with respect to x, we get:
\(\frac{d y}{d x}=\frac{d}{d x}\left(\sqrt{a^{2}-x^{2}}\right) \)
\(\Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{a^{2}-x^{2}}} \cdot \frac{d}{d x}\left(a^{2}-x^{2}\right) \)
\(=\frac{1}{2 \sqrt{a^{2}-x^{2}}}(-2 x) \)
\(=\frac{-x}{\sqrt{a^{2}-x^{2}}}\)
Substituting the value of \( \frac{d y}{d x}\) in the given differential equation, we get:
\(\text { L.H.S. } =x+y \frac{d y}{d x} =x+\sqrt{a^{2}-x^{2}} \times \frac{-x}{\sqrt{a^{2}-x^{2}}} \)
\(=x-x =0 =\text { R.H.S }\)
Hence, the given function is the solution of the corresponding differential equation. \(\)

The given equation can be rewritten as
\(x \sqrt{1-y^{2}} d x=-y \sqrt{1-x^{2}} d y \Rightarrow \frac{d y}{d x}=\frac{-x \sqrt{1-y^{2}}}{y \sqrt{1-x^{2}}} \)
Since, the highest order derivative is \( \frac{d y}{d x}.\)is 1. So, its degree is 1.

\(\int \frac{e^{\tan ^{4} x}}{1+x^{2}} d x=\int e^{t} d t=e^{t}+C=e^{\tan ^{-1} x}+C \)

Given differential equation is
\(y d x+\left(x-y^{2}\right) d y=0\) 
\(\Rightarrow y \frac{d x}{d y}+\left(x-y^{2}\right)=0\) 
\(\Rightarrow y \frac{d x}{d y}+x=y^{2}\) 
\(\Rightarrow \frac{d x}{d y}+\frac{1}{y} x=y\) [dividing both sides by y ]
which is of the form \(\frac{d x}{d y}+P x=Q\) 
Here,\(P=\frac{1}{y} \text { and } Q=y\) 
\(\therefore\) Integrating Factor \(\mathrm{IF}=e^{\int P d y}=e^{\int \frac{1}{y} d y}\) 
\(=e^{\log y}=y \quad\left[\because e^{\log f(x)}=f(x)\right]\)

\(I= \int { \sqrt { 1-4x-{ x }^{ 2 } } } dx\)
\(=\int { \sqrt { 1-({ x }^{ 2 }+4x) } } dx\)
\(=\int { \sqrt { 1+4-({ x }^{ 2 }+4x+x) } } dx\)
\(=\int { \sqrt { 5-{ (x+2) }^{ 2 } } } dx.\)
Put x+2=t so that dx=dt
\(\therefore I=\int { \sqrt { 5-{ t }^{ 2 } } } dt\)
\(=\int { \sqrt { 5-{ t }^{ 2 } } } dt \left[ From:\int { \sqrt { { a }^{ 2 }-{ x }^{ 2 } } dx } \right] \)
\(=\frac { t\sqrt { 5-{ t }^{ 2 } } }{ 2 } +\frac { 5 }{ 2 } \sin ^{ -1 }{ \frac { t }{ \sqrt { 5 } } } +C\)
\( =\frac { (x+2)\sqrt { 5-{ (x+2) }^{ 2 } } }{ 2 } +\frac { 5 }{ 2 } \sin ^{ -1 }{ \frac { x+2 }{ \sqrt { 5 } } } +C\)
\(=\frac { (x+2) }{ 2 } \sqrt { 1-4x-{ x }^{ 2 } } +\frac { 5 }{ 2 } \sin ^{ -1 }{ \left( \frac { x+2 }{ \sqrt { 5 } } \right) } +C\)

\(y =0 \Rightarrow x=0,4 \)
\(\text { Area } =\int_{0}^{4}\left(4 x-x^{2}\right) d x \)
\(\text { Area } =\left[2 x^{2}-\frac{x^{3}}{3}\right]_{0}^{4}=\left(32-\frac{64}{3}\right) \text { sq units } \)
\(=\frac{32}{3} \text { sq units } \)

(i) We have x² – 6x + 13 = x ²– 6x + 3² – 3² + 13 = (x – 3)² + 4
\(\text { So, } \ \int \frac{d x}{x^{2}-6 x+13}=\int \frac{1}{(x-3)^{2}+2^{2}} d x\)
Let x – 3 = t. Then dx = dt
\(\int \frac{d x}{x^{2}-6 x+13}=\int \frac{d t}{t^{2}+2^{2}}=\frac{1}{2} \tan ^{-1} \frac{t}{2}+C\)
\(=\frac{1}{2} \tan ^{-1} \frac{x-3}{2}+\mathrm{C}\)
(ii) The given integral is of the form 7.4 (7). We write the denominator of the integrand,
\(3 x^{2}+13 x-10=3\left(x^{2}+\frac{13 x}{3}-\frac{10}{3}\right)\)
\(=3\left[\left(x+\frac{13}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}\right] \text { (completing the square) }\)
\(\text { Thus } \int \frac{d x}{3 x^{2}+13 x-10}=\frac{1}{3} \int \frac{d x}{\left(x+\frac{13}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}} \)
\(\text { Put } x+\frac{13}{6}=t \text { . Then } d x=d t . \)
\(\text { Therefore, } \quad \int \frac{d x}{3 x^{2}+13 x-10}=\frac{1}{3} \int \frac{d t}{t^{2}-\left(\frac{17}{6}\right)^{2}} \)
\(=\frac{1}{3 \times 2 \times \frac{17}{6}} \log \left|\frac{t-\frac{17}{6}}{t+\frac{17}{6}}\right|+C_{1} \)
\(=\frac{1}{17} \log \left|\frac{x+\frac{13}{6}-\frac{17}{6}}{x+\frac{13}{6}+\frac{17}{6}}\right|+C_{1} \)
\(=\frac{1}{17} \log \left|\frac{6 x-4}{6 x+30}\right|+C_{1} \)
\(=\frac{1}{17} \log \left|\frac{3 x-2}{x+5}\right|+C_{1}+\frac{1}{17} \log \frac{1}{3} \)
\(=\frac{1}{17} \log \left|\frac{3 x-2}{x+5}\right|+C, \text { where } C=C_{1}+\frac{1}{17} \log \frac{1}{3} \)
\(\text { (iii) We have } \int \frac{d x}{\sqrt{5 x^{2}-2 x}}=\int \frac{d x}{\sqrt{5\left(x^{2}-\frac{2 x}{5}\right)}}\)
\(=\frac{1}{\sqrt{5}} \int \frac{d x}{\sqrt{\left(x-\frac{1}{5}\right)^{2}-\left(\frac{1}{5}\right)^{2}}} \text { (completing the square) }\)
\(\text { Put } x-\frac{1}{5}=t . \text { Then } d x=d t \text { . }\)
\(\text { Therefore, } \quad \int \frac{d x}{\sqrt{5 x^{2}-2 x}}=\frac{1}{\sqrt{5}} \int \frac{d t}{\sqrt{t^{2}-\left(\frac{1}{5}\right)^{2}}}\)
\(=\frac{1}{\sqrt{5}} \log \left|t+\sqrt{t^{2}-\left(\frac{1}{5}\right)^{2}}\right|+\mathrm{C}\)
\(=\frac{1}{\sqrt{5}} \log \left|x-\frac{1}{5}+\sqrt{x^{2}-\frac{2 x}{5}}\right|+\mathrm{C}\)

Given, \({ x }^{ 2 }dy+\left( xy+{ y }^{ 2 } \right) dx=0\)
\(\therefore \ \frac { dy }{ dx } =\frac { -\left( xy+{ y }^{ 2 } \right) }{ { x }^{ 2 } } \)
Put  y = vx
\(\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(\therefore \) The differential equation becomes
\(v+x\frac { dv }{ dx } =-\left( v+{ v }^{ 2 } \right) \)
\(\Rightarrow \frac { dv }{ { v }^{ 2 }+2v } =\frac { dx }{ x } \)
\(\Rightarrow \int { \frac { dv }{ \left( v+1 \right) ^{ 2 }-{ 1 }^{ 2 } } } =-\int { \frac { dx }{ x } } \)
\(\Rightarrow \frac { 1 }{ 2 } \log { \frac { v }{ v+2 } } =-\log { x } +\log { C } \)
\(\Rightarrow \frac { C }{ x } =\sqrt { \frac { y }{ y+2x } } \)
If x = 1, y = 1 then \(c=\frac { 1 }{ \sqrt { 3 } } \)
\(\Rightarrow \frac { 1 }{ \sqrt { 3 } x } =\sqrt { \frac { y }{ y+2x } } \)

As area is above the x-axis
∴ area = \(2\int _{ 0 }^{ 7 }{ \sqrt { 49-{ x }^{ 2 } } } \)
= \({ \left[ \frac { x }{ 2 } \sqrt { 49-{ x }^{ 2 } } +\frac { 49 }{ 2 } { sin }^{ -1 }\frac { x }{ 7 } \right] }_{ 0 }^{ 7 }\)
= \(2\left[ \left( \frac { 7 }{ 2 } \times 0+\frac { 49 }{ 2 } { sin }^{ -1 }1 \right) -(0) \right] \)
= \(\frac { 49 }{ 2 } \pi \) sq units

(i) Given, differential equation is (x² - y²)dx + 2xydy = 0
\(\begin{aligned} \Rightarrow \frac{d y}{d x} & =\frac{-\left(x^2-y^2\right)}{2 x y}=\frac{y^2-x^2}{2 x y} \end{aligned}\)
\(\begin{aligned} =\frac{x^2\left(\frac{y^2}{x^2}-1\right)}{2 x y}=\frac{\left(\frac{y}{x}\right)^2-1}{2\left(\frac{y}{x}\right)} \end{aligned}\)
\(\therefore\)  In RHS, degree of numerator and denominator is same
\(\therefore\)  It is a homogeneous differential equation and can be written as
\(\frac{d y}{d x}=g\left(\frac{y}{x}\right)\)
(ii) Given, differential equation is (x² - y²)dx + 2xy dy = 0
\(\Rightarrow \quad \frac{d y}{d x}=-\frac{\left(x^2-y^2\right)}{2 x y}=\frac{y^2-x^2}{2 x y}\) ...(i)
This is a homogeneous differential equation
On putting y = vx \(\Rightarrow \frac{d y}{d x}=v+x \cdot \frac{d v}{d x}\)
\(\therefore\) From Eq (i), we get
\(\begin{aligned} v+x \cdot \frac{d v}{d x} & =\frac{v^2-1}{2 v} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad x \frac{d v}{d x} & =\frac{v^2-1}{2 v}-v \end{aligned}\)
\(\begin{aligned} =\frac{v^2-1-2 v^2}{2 v}=\frac{-v^2-1}{2 v} \end{aligned}\)
\(\Rightarrow \frac{2 v}{v^2+1} d v=\frac{-d x}{x}\)
on integrating both sides, we get
log |v² + 1| = -log x + log c
\(\begin{aligned} & \Rightarrow \quad \log \left|\frac{y^2}{x^2}+1\right|=-\log x+\log c \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad \log \left|\frac{y^2+x^2}{x^2} \cdot x\right|=\log c \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad \frac{y^2+x^2}{x}=c \end{aligned}\)
\(\Rightarrow\)   y² + x² = cx
which is the required solution.

(a) Both A and R are correct; R is the correct explanation of A