Class 12th

$H:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Foci : S (ae, 0), S' (ae, 0)

Focus of parabola is (ae, 0)

Now, semi latus rectum of parabola = |SS'| = 2ae

$Given,4ae=e\left(\frac{2b^2}{a}\right)$

B² = 2a²……… (i)

$Given,\left(2\sqrt[]{2},-2\sqrt[]{2}\right)$ lies on H

$\Rightarrow \frac{1}{a^2}-\frac{1}{b^2}=\frac{1}{8}........(ii)$

From (i) and (ii)

$a^2=4,b^2=8$

$?b^2=a^2\left(e^2-1\right)$

$\therefore e=\sqrt[]{3}$

$\Rightarrow$ equation of parabola is  y² $=8\sqrt[]{3}x$

C : (x – 2)² + y² = 1

Equation of chord AB : 2x = 3

$OA=OB=\sqrt[]{3}$

$AM=\frac{\sqrt[]{3}}{2}$

$Areaof\Delta OAB=\frac{1}{2}\left(2AM\right)\left(OM\right)$

$=\frac{3\sqrt[]{3}}{4}sq.unit$

Equation of circle passing through (0, 2) and (0, 2) is

$x^2+\left(y^2-4\right)+\lambda x=0,\left(\lambda \in R\right)$

Divided by x we get

$\frac{x^2+\left(y^2-4\right)}{x}+\lambda =0$

Differentiating w.r.t. x

$\frac{x\left[2x+2y.\frac{dy}{dx}\right]-\left[x^2+y^2-4\right].1}{x^2}=0$

$\Rightarrow 2xy.\frac{dy}{dx}+\left(x^2-y^2+4\right)=0$

 $\frac{dy}{dx}+\frac{1}{x^2-1}y={\left(\frac{x-1}{x+1}\right)}^{1/2}$

$\frac{dy}{dx}+py=Q$

$I.F=e^{{\displaystyle \int Pdx}}={\left(\frac{x-1}{x+}\right)}^{\frac{1}{2}}$

$x-2lo{g}_e\left|x+1\right|+C$

Curves passes through $\left(2,\frac{1}{\sqrt[]{3}}\right)$

$\Rightarrow C=2lo{g}_{e}3-\frac{5}{3}$

$atx=8,\sqrt[]{7}y\left(8\right)=19-6lo{g}_{e}3$

Required area is

= ${\displaystyle \underset{e-e^2}{\overset{0}{\int }}\mathcal{l}n\left(x+e^2\right)-1dx+{\displaystyle \underset{0}{\overset{\mathcal{l}n2}{\int }}2e^{-x}-1dx}}$

$=1+e-\mathcal{l}n2$

 $l_n\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}\frac{dt}{{\left(t^2+s\right)}^n}}$

Applying integral by parts

$l_n\left(x\right)={\left[\frac{t}{{\left(t^2+5\right)}^n}\right]}_0^x-{\displaystyle \underset{0}{\overset{x}{\int }}n{\left(t^2+5\right)}^{-n-1}.2t^2}$

$10n{l}_{n+1}\left(x\right)+\left(1-2n\right)l_n\left(x\right)=\frac{x}{{\left(x^2+5\right)}^n}$

Put n = 5

 $x=2\sqrt[]{2}cost\sqrt[]{sin2t}$

$\frac{dx}{dt}=\frac{2\sqrt[]{2}cos3t}{\sqrt[]{sin2t}}$

$y\left(t\right)=2\sqrt[]{2}sint\sqrt[]{sin2t}$

$\frac{dy}{dx}=\frac{2\sqrt[]{2}sin3t}{\sqrt[]{sin2t}}$

$\therefore \frac{1+{\left(\frac{dy}{dx}\right)}^2}{\frac{d^{2}y}{d{x}^2}}=\frac{1+1}{-3}=\frac{-2}{3}$

 $f\left(x\right)=ta{n}^{-1}\left(sinx-cosx\right)$

$f\text{'}\left(x\right)=\frac{cosx+sinx}{{\left(sinx-cosx\right)}^2+1}$ = 0

$\therefore x=\frac{3\pi }{4}$

Sum = tan^-1 $\sqrt[]{2}-\frac{\pi }{4}$

= $co{s}^{-1}\frac{1}{\sqrt[]{3}}-\frac{\pi }{4}$

 $f\left(x\right)=x{e}^{x\left(1-x\right)}$

$f\text{'}\left(x\right)=-e^{x\left(1-x\right)}\left(2x+1\right)\left(x-1\right)$

$f\left(x\right)is\uparrow in\left(\frac{-1}{2},-1\right)$

Note : n should be given as a natural number:

$f\left(x\right)=\{\begin{array}{l}\begin{array}{ccc}\hfill \frac{-sin\left(x-1\right)}{x-1}\hfill & \hfill ,\hfill & \hfill x<-1\hfill \\ \hfill -\left(sin2+1\right)\hfill & \hfill ,\hfill & \hfill x=-1\hfill \\ \hfill cos2\pi x\hfill & \hfill ,\hfill & \hfill -1<x<1\hfill \end{array}\\ 1,x=1\end{array}$

$\frac{-sin\left(x-1\right)}{x-1},x>1$

f (x) is discontinuous at x = 1 and x = 1