# Current Affairs JEE Main & Advanced

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

Category : JEE Main & Advanced

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.

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