# Current Affairs JEE Main & Advanced

#### Distance Between The Pair of Parallel Straight Lines

If $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ represents a pair of parallel straight lines, then the distance between them is given by $2\sqrt{\frac{{{g}^{2}}-ac}{a(a+b)}}$or $2\sqrt{\frac{{{f}^{2}}-bc}{b(a+b)}}$.

#### Removal of the term $\mathbf{xy}$ from $\mathbf{f(X,}\,\mathbf{Y)=a}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+2hxy+b}{{\mathbf{y}}^{\mathbf{2}}}$ Without Changing the Origin

Clearly, $h\ne 0$. Rotating the axes through an angle $\theta$, we have, $x=X\cos \theta -Y\sin \theta$ and  $y=X\,\sin \theta +Y\cos \theta$   $\therefore$   $f(x,y)=a{{x}^{2}}+2hxy+b{{y}^{2}}$   After rotation, new equation is   $F(X,Y)=(a{{\cos }^{2}}\theta +2h\cos \theta \sin \theta +b{{\sin }^{2}}\text{ }\theta ){{X}^{2}}$   $+2\{(b-a)\cos \theta \sin \theta +h({{\cos }^{2}}\theta -{{\sin }^{2}}\theta )XY$      $+(a{{\sin }^{2}}\theta -2h\cos \theta \sin \theta +b{{\cos }^{2}}\theta ){{Y}^{2}}$     Now coefficient of $XY=0$. Then we get $\cot 2\theta =\frac{a-b}{2h}$.     Usually, we use the formula, $\tan 2\theta =\frac{2h}{a-b}$ for finding the angle of rotation $\theta$. However, if $a=b$, we use $\cot 2\theta =\frac{a-b}{2h}$ as in this case $\tan 2\theta$ is not defined.

#### Removal of First Degree Terms

Let point of intersection of lines represented by $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$       .....(i)  is $(\alpha ,\beta )$.   Here $(\alpha ,\beta )=\left( \frac{bg-fh}{{{h}^{2}}-ab},\frac{af-gh}{{{h}^{2}}-ab} \right)$   For removal of first degree terms, shift the origin to $(\alpha ,\beta )$ i.e., replacing $x$ by $(X+\alpha )$and $y$ be $(Y+\beta )$in (i).   Alternative Method : Direct equation after removal of first degree terms is $a{{X}^{2}}+2hXY+b{{Y}^{2}}+(g\alpha +f\beta +c)=0$,   where $\alpha =\frac{bg-fh}{{{h}^{2}}-ab}$ and $\beta =\frac{af-gh}{{{h}^{2}}-ab}$.

#### Equation of the Lines Joining the Origin to the Points of Intersection of a Given Line and a Given Curve

The equation of the lines which joins origin to the point of intersection of the line $lx+my+n=0$ and curve $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$, can be obtained by making the curve homogeneous with the help of line $lx+my+n=0$, which is         $a{{x}^{2}}+2hxy+b{{y}^{2}}+2(gx+fy)\left( \frac{lx+my}{-n} \right)+c\,{{\left( \frac{lx+my}{-n} \right)}^{2}}=0$

#### Bisectors of the Angles Between the Lines

(1) The joint equation of the bisectors of the angles between the lines represented by the equation $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ is   $\frac{{{x}^{2}}-{{y}^{2}}}{a-b}=\frac{xy}{h}\Rightarrow h{{x}^{2}}-(a-b)xy-h{{y}^{2}}=0$   Here, coefficient of ${{x}^{2}}+$ coefficient of ${{y}^{2}}=0$. Hence, the bisectors of the angles between the lines are perpendicular to each other. The bisector lines will pass through origin also.   (i) If $a=b$, the bisectors are ${{x}^{2}}-{{y}^{2}}=0$.   i.e., $x-y=0,x+y=0$   (ii) If $h=0$, the bisectors are $xy=0$ i.e., $x=0,y=0$.   (2) The equation of the bisectors of the angles between the lines represented by  $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ are given by $\frac{{{(x-\alpha )}^{2}}-{{(y-\beta )}^{2}}}{a-b}=\frac{(x-\alpha )(y-\beta )}{h}$, where $\alpha ,\,\,\beta$ is the point of intersection of the lines represented by the given equation.

#### Angle Between the Pair of Lines

The angle between the lines represented by        $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ or $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$     is given by $\tan \theta =\left| \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right|\,\,\,\,\Rightarrow \theta ={{\tan }^{-1}}\left| \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right|$     From the above formula it is clear, that     (i) The lines represented by $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ are parallel iff ${{h}^{2}}=ab$ and $a{{f}^{2}}=b{{g}^{2}}$ or $\frac{a}{h}=\frac{h}{b}=\frac{g}{f}$.     (ii) The lines represented by $a{{x}^{2}}+2hxy+b{{y}^{2}}$ $+2gx+2fy+c=0$ are perpendicular iff $a+b=0$     i.e.,  Coefficient of ${{x}^{2}}+$ Coefficient of ${{y}^{2}}=0$.     (iii) The lines are coincident, if ${{g}^{2}}=ac$.

#### Point of Intersection of Lines Represented by $\mathbf{a}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+ 2hxy+b}{{\mathbf{y}}^{\mathbf{2}}}\mathbf{+ 2gx+ 2fy+c= 0}$

Let $\varphi \equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$     $\frac{\partial \varphi }{\partial x}=2ax+2hy+2g=0$                          (Keeping $y$ as constant)       and $\frac{\partial \varphi }{\partial y}=2hx+2by+2f=0$    (Keeping $x$ as constant)     For point of intersection $\frac{\partial \varphi }{\partial x}=0$ and $\frac{\partial \varphi }{\partial y}=0$     We obtain, $ax+hy+g=0$ and $hx+by+f=0$     On solving these equations, we get     $\frac{x}{fh-bg}=\frac{y}{gh-af}=\frac{1}{ab-{{h}^{2}}}$i.e.,$(x,y)=\left( \frac{bg-fh}{{{h}^{2}}-ab},\frac{af-gh}{{{h}^{2}}-ab} \right)$.     (3) Separate equations from joint equation: The general equation of second degree be $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$     To find the lines represented by this equation we proceed as follows :     Step I : Factorize the homogeneous part $a{{x}^{2}}+2hxy+b{{y}^{2}}$ into two linear factors. Let the linear factors be $a'x+b'y$ and $a''x+b''y$.     Step II : Add constants $c'$and $c''$ in the factors obtained in step I to obtain $a'x+b'y+c'$ and $a''x+b''y+c''$. Let the lines be $a'x+b'y+c'=0$ and $a''x+b''y+c''=0$.     Step III : Obtain the joint equation of the lines in step II and compare the coefficients of $x,\,\,y$ and constant terms to obtain equations in $c'$and $c''$.     Step IV : Solve the equations in $c'$and $c''$ to obtain the values of $c'$and $c''$.     Step V : Substitute the values of $c'$and $c''$ in lines in step II to obtain the required lines.

#### Equation of Pair of Straight Lines

(1) Equation of a pair of straight lines passing through origin : The equation $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents a pair of straight line passing through the origin where $a,\,\,h,\,\,b$ are constants.   Let the lines represented by $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ be $y-{{m}_{1}}x=0,\,\,y-{{m}_{2}}x=0$. Then, ${{m}_{1}}+{{m}_{2}}=-\frac{2h}{b}$ and ${{m}_{1}}{{m}_{2}}=\frac{a}{b}$   Then, two straight lines represented by $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ are $ax+hy+y\sqrt{{{h}^{2}}-ab}=0$ and $ax+hy-y\sqrt{{{h}^{2}}-ab}=0$.     Hence, (a) The lines are real and distinct, if ${{h}^{2}}-ab>0$     (b) The lines are real and coincident, if ${{h}^{2}}-ab=0$     (c) The lines are imaginary, if ${{h}^{2}}-ab<0$     (2) General equation of a pair of straight lines : An equation of the form, $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ where $a,\,\,b,\,\,c,\,\,f,\,\,g,\,\,h$ are constants, is said to be a general equation of second degree in $x$ and $y$.     The necessary and sufficient condition for $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ to represents a pair of straight lines is that $abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$ or $\left| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \\ \end{matrix} \right|=0$.

#### Logic Gates

(i) AND : It is the boolean function defined by   $f({{x}_{1}},{{x}_{2}})={{x}_{1}}\wedge {{x}_{2}}$; ${{x}_{1}},\,{{x}_{2}}\in \{0,\,1\}$.   It is shown in the figure given below.
 Input Output ${{x}_{1}}$ ${{x}_{2}}$ ${{x}_{1}}\wedge {{x}_{2}}$ 1 1 0 0 1 0 1 0 1 0 0 0
(ii) OR : It is the boolean function defined by   $f({{x}_{1}},{{x}_{2}})={{x}_{1}}\vee {{x}_{2}}$; ${{x}_{1}},{{x}_{2}}\in \{0,\,1\}$.   It is shown in the figure given below
 Input Output ${{x}_{1}}$ ${{x}_{2}}$ ${{x}_{1}}\vee {{x}_{2}}$ 1 1 0 0 1 0 1 0 1 1 1 0
(iii) NOT : It is the boolean function defined by   $f(x)={x}',$ $x\in \{0.1\}$   It is shown in the figure given below:         more...

#### Switching Circuits

One of the major practical application of Boolean algebra is to the switching systems (an electrical network consisting of switches) that involves two state devices. The simplest possible example of such a device is an ordinary ON-OFF switch.   By a switch we mean a contact or a device in an electric circuit which lets (or does not let) the current to flow through the circuit. The switch can assume two states ‘closed’ or ‘open’ (ON or OFF). In the first case the current flows and in the second the current does not flow.   Symbols $a,\,b,\,c,\,p,\,q,\,r,\,x,y,\,z$,..... etc. will denote switches in a circuit.   There are two basic ways in which switches are generally interconnected.   (i) Series   (ii) Parallel     (i) Series : Two switches a, b are said to be connected ‘in series’ if the current can pass only when both are in closed state and the current does not flow if any one or both are open. The following diagram will show this circuit.      (ii) Parallel : Two switches $a,b$ are said to be connected ?in parallel? if current flows when any one or both are closed, and current does not pass when both are open. The following diagram will represent this circuit given by $a\vee b$.   If two switches in a circuit be such that both are open (closed) simultaneously, we shall represent them by the same letter. Again if two switches be such that one is open iff the other is closed, we represent them by a and a¢.   The value of a close switch or when it is on is equal to 1 and when it is open or off is equal to 0.   An open switch r is indicated in the diagram as follows :       A closed switch r is indicated in the diagram as follows :       Boolean operations on switching circuits   (i) Boolean Multiplication : The two switches r and s in the series will perform the operation of Boolean multiplication.        Clearly, the current will not pass from point ${{S}_{1}}$ to ${{S}_{2}}$ when either or both r, s are open. It will pass only when both are closed.
r s more...

#### Trending Current Affairs

You need to login to perform this action.
You will be redirected in 3 sec