Class 12th

20.  (iv) They are anisotropic in nature since amorphous solid shows isotropic properties as they exhibit same values of the properties like refractive index, electrical resistance when measured along different directions.

The equation of a circle in the first quadrant with centre (a, a) and radius (a) which touches the coordinate axes is:

${(x-a)}^2+{(y-a)}^2=a^2..........(1)$

Differentiating equation (1) with respect to x, we get:

$\begin{array}{l}2(x-a)+2(y-a)\frac{dy}{dx}=0\\ \Rightarrow (x-a)+(y-a)y^{\text{'}}=0\\ \Rightarrow x-a+y{y}^{\text{'}}-a{y}^{\text{'}}=0\\ \Rightarrow x+y{y}^{\text{'}}-a(1+y^{\text{'}})=0\\ \Rightarrow a=\frac{x+y{y}^{\text{'}}}{1+y^{\text{'}}}\end{array}$

Substituting the value of a in equation (1), we get:

$\begin{array}{l}{\left[x-\left(\frac{x+y{y}^{\text{'}}}{1+y^{\text{'}}}\right)\right]}^2+{\left[y-\left(\frac{x+y{y}^{\text{'}}}{1+y^{\text{'}}}\right)\right]}^2={\left(\frac{x+y{y}^{\text{'}}}{1+y^{\text{'}}}\right)}^2\\ \Rightarrow {\left[\frac{(x-a)y^{\text{'}}}{(1+y^{\text{'}})}\right]}^2+{\left[\frac{y-x}{1+y^{\text{'}}}\right]}^2={\left[\frac{x+y{y}^{\text{'}}}{1+y^{\text{'}}}\right]}^2\\ \Rightarrow {(x-y)}^2.y^{\text{'}2}+{(x-y)}^2=(x+y{y}^{\text{'}}){}^2\\ \Rightarrow {(x-y)}^2\left[1+(y^{\text{'}}){}^2\right]=(x+y{y}^{\text{'}}){}^2\end{array}$

Hence, the required differential equation of the family of circles is ${(x-y)}^2\left[1+(y^{\text{'}}){}^2\right]=(x+y{y}^{\text{'}}){}^2$

$\Rightarrow \frac{dy}{dx}=\frac{x^3-3x{y}^2}{y^3-3x^{2}y}..........(1)$

This is a homogenous equation. To simplify it, we need to make the substitution as:

$\begin{array}{l}y=vx\\ \Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(vx)\\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\end{array}$

Substituting the values of y and $\frac{dv}{dx}$ in equation (1), we get:

$\begin{array}{l}v+x\frac{dv}{dx}=\frac{x^3-3x{(vx)}^2}{{(vx)}^3-3x^2(vx)}\\ \Rightarrow v+x\frac{dv}{dx}=\frac{1-3v^2}{v^3-3v}\\ \Rightarrow x\frac{dv}{dx}=\frac{1-3v^2}{v^3-3v}-v\\ \Rightarrow x\frac{dv}{dx}=\frac{1-3v^2-v(v^3-3v)}{v^3-3v}\\ \Rightarrow x\frac{dv}{dx}=\frac{1-v^4}{v^3-3v}\\ \Rightarrow \left(\frac{v^3-3v}{1-v^4}\right)dv=\frac{dx}{x}\end{array}$

Integrating both sides, we get:

$\begin{array}{l}{\displaystyle \int \left(\frac{v^3-3v}{1-v^4}\right)dv=logx+log{C}^{\text{'}}}.........(2)\\ Now,{\displaystyle \int \left(\frac{v^3-3v}{1-v^4}\right)dv={\displaystyle \int \frac{v^{3}dv}{1-v^4}}}-3{\displaystyle \int \frac{vdv}{1-v^4}}\\ \Rightarrow {\displaystyle \int \left(\frac{v^3-3v}{1-v^4}\right)}dv=I_1-3I_2,Where,I_1={\displaystyle \int \frac{v^{3}dv}{1-v^4}}and{I}_2={\displaystyle \int \frac{vdv}{1-v^4}...........(3)}\end{array}$

$\begin{array}{l}Let,1-v^4=t.\\ \therefore \frac{d}{dv}(1-v^4)=\frac{dt}{dv}\\ \Rightarrow -4v^3=\frac{dt}{dv}\\ \Rightarrow v^{3}dv=-\frac{dt}{4}\\ Now,I_1={\displaystyle \int -\frac{dt}{4}=-logt=-\frac{1}{4}}log(1-v^4)\end{array}$

$\begin{array}{l}And,I_2={\displaystyle \int \frac{vdv}{1-v^4}}={\displaystyle \int \frac{vdv}{1-(v^2){}^2}}\\ Let,v^2=p.\\ \therefore \frac{d}{dv}\left(v^2\right)=\frac{dp}{dv}\\ \Rightarrow 2v=\frac{dp}{dv}\\ \Rightarrow vdv=\frac{p}{2}\\ \Rightarrow I_2=\frac{1}{2}{\displaystyle \int \frac{dp}{1-p^2}=\frac{1}{2*2}log\left|\frac{1+p}{1-p}\right|=\frac{1}{4}log}\left|\frac{1+v^2}{1-v^2}\right|\end{array}$

Substituting the values of $I_1$ and $I_2$ in equation (3), we get:

${\displaystyle \int \left(\frac{v^3-3v}{1-v^4}\right)}dv=-\frac{1}{4}log(1-v^4)-\frac{3}{4}log\left|\frac{1+v^2}{1-v^2}\right|$

Therefore, equation (2) becomes:

$\begin{array}{l}\frac{1}{4}log(1-v^4)-\frac{3}{4}log\left|\frac{1+v^2}{1-v^2}\right|=logx+log{C}^{\text{'}}\\ \Rightarrow -\frac{1}{4}log\left[(1-v^4)\left(\frac{1+v^2}{1-v^2}\right)\right]=log{C}^{\text{'}}x\\ \Rightarrow \frac{{\left(1+v^2\right)}^4}{{\left(1-v^2\right)}^2}={\left(C^{\text{'}}x\right)}^{-4}\\ \Rightarrow \frac{{\left(1+\frac{y^2}{x^2}\right)}^4}{{\left(1-\frac{y^2}{x^2}\right)}^2}=\frac{1}{C^{\text{'}4}{x}^4}\\ \Rightarrow \frac{{\left(x^2+y^2\right)}^4}{x^4{\left(x^2-y^2\right)}^2}=\frac{1}{C^{\text{'}4}{x}^4}\\ \Rightarrow {\left(x^2-y^2\right)}^2=C^{\text{'}4}{\left(x^2+y^2\right)}^4\\ \Rightarrow \left(x^2-y^2\right)=C^{\text{'}2}{\left(x^2+y^2\right)}^2\\ \Rightarrow x^2-y^2=C{\left(x^2+y^2\right)}^2,whereC=C^{\text{'}2}\end{array}$

Hence, the given result is proved.

Equation of the given family of curves is  ${(x-a)}^2+2y^2=a^2$

$\begin{array}{l}{(x-a)}^2+2y^2=a^2\\ \Rightarrow x^2+a^2-2ax+2y^2=a^2\\ \Rightarrow 2y^2=2ax-x^2..........(1)\end{array}$

Differentiating with respect to x, we get:

$\begin{array}{l}2y\frac{dy}{dx}=\frac{2a-2x}{2}\\ \Rightarrow \frac{dy}{dx}=\frac{a-x}{2y}\\ \Rightarrow \frac{dy}{dx}=\frac{2a-2x^2}{4xy}..........(2)\end{array}$

From equation (*1), we get:

$2ax=2y^2+x^2$

On substituting this value in equation (3), we get:

$\begin{array}{l}\frac{dy}{dx}=\frac{2y^2+x^2-2x^2}{4xy}\\ \Rightarrow \frac{dy}{dx}=\frac{2y^2-x^2}{4xy}\end{array}$

Hence, the differential equation of the family of curves is given as $\frac{dy}{dx}=\frac{2y^2-x^2}{4xy}$

19. Option (i) Same in all directions is correct since quartz glass is an amorphous solid showing isotropic properties and hence exhibits the same values of refractive index when measured along different directions.