Category : 11th Class
Pair and Straight Line
Key Points to Remember
A hemogenous of equation of second degree of the form \[a{{x}^{2}}+2hxy+b{{y}^{2}}=0\] represents the pair of straight lines which passes through the origin.
(a) If the lines are real and distinct then
\[{{h}^{2}}>ab\]
(b) If the lines are real and coincidents if
\[{{h}^{2}}=ab.\]
(c) If the lines are imaginary then \[{{h}^{2}}<ab.\]
Let \[y={{m}_{1}}x\to (1)\] and \[y={{m}_{2}}x\to (2)\] are two lines which are passing through the origin.
Then \[(y-{{m}_{1}}x)(y-{{m}_{2}}x)\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}\]
\[{{y}^{2}}-({{m}_{1}}+{{m}_{2}})xy+{{m}_{1}}{{m}_{2}}{{x}^{2}}={{y}^{2}}+\frac{2h}{b}xy+\frac{a}{b}{{x}^{2}}\]
Equation the coefficient of same variable of the\[\frac{2h}{b}=({{m}_{1}}+{{m}_{2}})\]we have, \[({{m}_{1}}+{{m}_{2}})=\frac{2h}{b}\] & \[{{m}_{1}}.{{m}_{2}}=\frac{a}{b}\]
Let q be the angle between the two given pair of straight line \[a{{x}^{2}}+2hxy+b{{y}^{2}}=0,\] which are passing through origin is written as
\[\tan \theta =\pm \frac{2\sqrt{{{h}^{2}}}-ab}{a+b}\]
for acute angle,
\[\tan \theta =\left| \frac{2\sqrt{{{h}^{2}}}-ab}{a+b} \right|\]
In cosine form
\[\cos \theta =\frac{a+b}{\sqrt{{{(a+b)}^{2}}+4{{h}^{2}}}}\]
Note: If two lines are coincident
i.e. \[\theta =0{}^\circ \] or \[180{}^\circ \]
then \[\frac{2\sqrt{{{h}^{2}}-ab}}{a+b}=0\]
Hence \[{{h}^{2}}=ab\Rightarrow {{h}^{2}}=a\]
(b) If two lines are perpendicular
i.e. \[\theta =90{}^\circ ,\] i.e. \[\tan 90{}^\circ =\infty \]
then Loosly, it can be written
\[\frac{2\sqrt{{{h}^{2}}-ab}}{a+b}=\infty =\frac{1}{0}\]
\[a+b=0\] i.e.
\[\therefore \]Sum of the coefficient of \[{{x}^{2}}\] and \[{{y}^{2}}\] respectively is zero.
(c) The equation of pair of straight lines passing through the origin and perpendicular to the given equation of pair of straight lines\[/a{{x}^{2}}+2hxy+b{{y}^{2}}=0\] is written as \[b{{x}^{2}}-2hxy+a{{x}^{2}}=0\]
The general equation of second degree in x and y be
\[a{{x}^{2}}+b{{y}^{2}}+2hxy+2gx+2fy+c=0\] …….(i)
represents a pair of straight lines iff
\[abc+2fgx-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0\]
(a) If \[\Delta \ne 0\] & \[{{h}^{2}}-ab\ge 0\] then this general equation of second degree in x & y represents the equation of hyperbola.
(b) If \[\Delta \ne 0\] & \[{{h}^{2}}-ab\le 0\] then this pair of lines represent the equation ellipse.
(c) If \[\Delta \ne 0\] & \[{{h}^{2}}-ab=0\] then this pair of lines represent the equation of parabola
(d) If \[a=b=1\] & \[h=0\] then this represents the equation of circle.
(a) Angle between the lines:-
If \[\theta \] is the angle between the two lines:
\[a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\] …… (1)
then \[\tan \theta =\pm \left| \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right|\]
\[\Rightarrow \,\,\,\theta ={{\tan }^{-1}}\left| \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right|\]
Note: It is the same form which is obtained by the pair of straight lines passing through the origin.
(b) Point of intersection of lines:
The point of intersection of line (1) is obtained by the partially differentiation of
\[\text{f}\equiv a{{x}^{\text{2}}}+b{{y}^{2}}+2hxy+2gx+2\text{f}y+c=0\]
w.r.t x and y respective & making
\[\frac{\partial \text{f}}{\partial x}=0\] ….. (i)
& \[\frac{\partial \text{f}}{\partial y}=0\] ….. (ii)
Then solve these equations and hence, we will obtain the value ofx and y, which are written as
\[(x,y)=\left( \frac{bg-h}{{{h}^{2}}-ab},\frac{a\text{f-gh}}{{{h}^{2}}-ab} \right)\]
Here,
\[\frac{\partial \text{f}}{\partial x}=2ax+2hy+2g=0\] & \[\frac{\partial \text{f}}{\partial x}=2by+2hx+2\text{f}=0\]
If the two lines represented by equation (1) be parallel if \[{{h}^{2}}=ab\] & \[b{{g}^{2}}=a{{\text{f}}^{\text{2}}}\]
\[d=2\sqrt{\frac{{{g}^{2}}-ac}{a(a+b)}}\]
The equation to the pair of lines through the origin and forming an equiletral triangle with line
\[ax+by+c=0\]is given by \[{{(ax+by)}^{2}}-3{{(ay-bx)}^{2}}=0\]
Also, the area of equiletral triangle is
\[=\frac{{{c}^{2}}}{\sqrt{3}({{a}^{2}}+{{b}^{2}})}\]
The line \[a{{x}^{2}}+b{{y}^{2}}+2hxy=0\] & \[\ell \times +my+n\]
= 0 'form issosectes triangle if
\[\frac{{{\ell }^{2}}-{{m}^{2}}}{\ell m}=\frac{a-b}{h}\]
Area of the triangle = \[\left| \frac{{{n}^{2}}\sqrt{{{h}^{2}}-ab}}{a{{m}^{2}}-2h\ell m+b{{\ell }^{2}}} \right|\]
\[\frac{{{x}^{2}}-{{y}^{2}}}{a-b}=\frac{xy}{h}\]
Let the straight line is
\[y=mx+c\]
\[\Rightarrow y-mx=c\Rightarrow \frac{y-mx}{c}=1\]
& The equation of the curve is
\[a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2\text{f}y+K=0\]
Let the line cuts the curve at point P & Q then the joint equation of OP & OQ be written as
\[a{{x}^{2}}+2hxy+b{{y}^{2}}+(2gx+2\text{f}y)+K{{\left( \frac{y-mx}{c} \right)}^{2}}=0\]
From this method we can make any pair of equation to be homogenous with the help of the equation of the line.
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