Homogeneous Equation

Homogeneous equation in two variable

Homogeneous equation in two variable.

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Homogeneous equation in two variable i.e $ax^2 + 2hxy +by^2 = 0$  always represent pair of line.

Since $g=f=c=0\;\Rightarrow bg^2+ch^2+af^2-2hfg-abc=0\,\forall\,x \in R$

Pair of lines represented by homogeneous equation in two variables are $L_1:y – m_1x = 0$ and $L_2 : y – m_2x = 0$, where $m_1,m_2$ are slopes of two lines that passes through origin.

Hence Equation of lines $L_1L_2=(y – m_1x)(y – m_2x) = 0$ and given that the pair of lines $L_1L_2 = ax^2+2hxy+by^2=0$.

Hence from both equations we get

$\cfrac{m_1m_2}{a}=\cfrac{- (m_1+m_2)}{2h}=\cfrac{1}{b}$

$m_1m_2=\cfrac{a}{b}$ and $m_1+m_2 = – \cfrac{2h}{b}$

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  Angle between lines represented by homogeneous equation in two variable $ax^2+2hxy+by^2=0 $ $$ $$

$m_1\,,\,m_2$ are slope of lines $L_1=0$ and $L_2=0$ respectively. Let $\theta$ be the angle between the lines then,

$tan\theta = \vert \cfrac{m_1 – m_2}{1 + m_1m_2} \vert = \vert \cfrac{2\sqrt{h^2 – ab}}{a + b} \vert\;\;since\; m_1+m_2= – \cfrac{2h}{b}\;and\;m_1m_2=\cfrac{a}{b}$

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Product of perpendiculars from a point $P(x_1 , y_1)$ on the pair of lines represented by $ax^2 + 2hxy + by^2 = 0$ $$ $$

$y = m_1x \;and \; y = m_2x$ be the two lines represented combinedly as $ax^2 + 2hxy + by^2 = 0$.

$\Rightarrow m_1 + m_2 = – \cfrac{2h}{b}\;and\; m_1m_2 = \cfrac{a}{b}$

Distance of point $P(x_1 , y_2)$ from line $y = mx$ is given by $P = \cfrac{\vert y_1 – mx_1 \vert}{\sqrt{1 + m_1^2}}$

Hence product of distance of point $P(x_1 , y_1)$ from lines is $y = m_1x\;and\;y = m_2x$ is given by

$P_1P_2 = \cfrac{\vert y_1 – m_1x_1 \vert}{\sqrt{1 + m_1^2}}\cfrac{\vert y_1 – m_2x_1 \vert}{\sqrt{1 + m_2^2}}$

$=\cfrac{\vert (y_1 – m_1x_1)(y_1 – m_2x_1) \vert}{\sqrt{1 + m_1^2}\sqrt{1 + m_2^2}}$

$= \cfrac{1}{\vert b \vert}\cfrac{\vert ax_1^2 + 2hx_1y_1 + by_1^2 \vert}{\sqrt{1 + m_1^2 + m_2^2 + m_1m_2}}$

$=\cfrac{1}{\vert b \vert}\cfrac{\vert ax_1^2 + 2hx_1y_1 + by_1^2 \vert}{\sqrt{(m_1 + m_2)^2 + (1 – m_1m_2)}}$

$=\cfrac{\vert ax_1^2 + 2hx_1y_1 + by_1^2 \vert}{\sqrt{(a – b)^2 + 4h^2}}$

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Angle bisector of lines given homogeneous equation in two variable $ax^2 + 2hxy + by^2 = 0$ $$ $$

$y = m_1x \;and \; y = m_2x$ be the two lines represented combinedly as $ax^2 + 2hxy + by^2 = 0$.

$\therefore$ Equation of bisectors are

$\cfrac{y – m_1x}{\sqrt{1 + m_1^2}} = \cfrac{y – m_2x}{\sqrt{1 + m_2^2}}\;—(1) $

$\cfrac{y – m_1x}{\sqrt{1 + m_1^2}} = – \cfrac{y – m_2x}{\sqrt{1 + m_2^2}}\;—(2) $

$\therefore$ Combined equation of bisector is

$\Bigl(\cfrac{y – m_1x}{\sqrt{1 + m_1^2}}\Bigr)^2 – \Bigl(\cfrac{y – m_2x}{\sqrt{1 + m_2^2}}\Bigr)^2 = 0$

$\therefore (1+m_2^2)(y^2+m_1^2x^2-2m_1xy) – (1+m_1^2)(y^2+m_2^2x^2-2m_2xy)=0$

$(m_1^2-m_2^2)y^2 + (m_1^2+\cancel{m_1^2m_2^2}-m_2^2-\cancel{m_1^2m_2^2})x^2-2(m_1+m_1m_2^2-m_2-m_1^2m_2)xy = 0$

$(m_2 – m_1) ((m_1+m_2)(y^2 – x^2) + 2(1 – m_1m_2)xy)=0$

Since $m_1\ne m_2\;\therefore -\cfrac{2h}{b}(y^2-x^2)-2\Bigl(\cfrac{b – a}{b}\Bigr)xy = 0$

$(x^2 – y^2)h = (a – b)xy$

$\cfrac{x^2 – y^2}{a – b} = \cfrac{xy}{h}$ $$ $$

Note :-          Let given pair of lines be $ax^2+2hxy+by^2+2gx+2fy+c=0$ intersect each other at point $A (\alpha , \beta)$ then equation angle bisector is given by

                        $\cfrac{(x – \alpha)^2 – (y – \beta)^2}{a – b} = \cfrac{(x – \alpha)(y – \beta)}{h}$

Concept of Homogenisation

Let $S(x,y)=ax^2 + 2hxy + by^2 + 2gx + 2fy + c =0$ represent one of the conic section, i.e circle, parabola, ellipse or hyperbola and the line $L=lx + my + n = 0$ intersect the curve at A and B.

Combined equation of OA and OB, where O is the origin is,

From the equation of a line $L=0$ we rewrite the equation as

$\cfrac{lx + my}{-n}=1\;——–(1)$

equation of curve $S(x,y)=0$ can be written as $ax^2 + 2hxy + by^2 + 2(gx + fy).1 + c.1^2=0\;—–(2)$

From (1) and (2) we get,

$ax^2 + 2hxy + by^2 + 2(gx + fy)\Bigl(\cfrac{lx + my}{-n}\Bigr) + c\Bigl(\cfrac{lx + my}{-n}\Bigr)^2=0$

The above equation represent equation of pair ol lines OA and OB.

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